Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
9th Edition
ISBN: 9781133949640
Author: John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher: Cengage Learning
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Textbook Question
Chapter 17, Problem 107IL

For the titration of 50.0 mL of 0.150 M ethylamine. C2H5NH2, with 0.100 M HCl, find the pH at each of the following points, and then use that information to sketch the titration curve and decide on an appropriate indicator.

  1. (a) At the beginning, before HCl is added
  2. (b) At the halfway point in the titration
  3. (c) When 75% of the required acid has been added
  4. (d) At the equivalence point
  5. (e) When 10.0 mL more HCl has been added than is required
  6. (f) Sketch the titration curve.
  7. (g) Suggest an appropriate indicator for this titration.

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The value of pH for the original solution of C2H5NH2 has to be calculated.

Concept introduction:

Titration is a quantitative method to determine the quantity of an acid or base in a solution. This method is used to determine the concentration an acid in the solution by titrating it against a base.

For strong acid-weak base titration the pH value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of C2H5NH2 with HCl is represented as,

C2H5NH2(aq)+ H3O+(aq)H2O(l)+C2H5NH3+(aq)

Calculation of pH at various points is done as follows,

(1) The pOH value before the titration can be calculated by using Kb and its relation with OH ion concentration.

Kb=[OH](eq)[BH+](eq)[B](eq) (1)

(2) The pH calculation just before the equivalence point,

As the addition of HCl is done there will be formation of buffer solution C2H5NH2/C2H5NH3+. The pOH calculation for buffer solution is done by using Henderson-Hasselbalch equation.

pOH=pKb+log[conjugate acid][base] (2)

At the midpoint of the titration, concentration of base and its conjugate acid is equal. Therefore pOH value at midpoint will be given as

pOH=pKb+log[conjugate acid][base]

Substitute, [conjugateacid]for[base].

pOH=pKb+log[base][base]=pKb+log(1)=pKb+0=pKb

Therefore, pOH value at midpoint is equal to pKb.

(3) The pOH calculation the equivalence point.

At equivalence point all the acid will be neutralized, and there will be only C2H5NH3+. The H3O+ will be produced due to the hydrolysis of conjugate acid at equivalence point. The hydrolysis equilibrium is represented as,

C2H5NH3+(aq)+H2O(l)H3O+(aq)+C2H5NH2(aq)

By using the value of Ka for the ethylamine, concentration of  H3O+ can be calculated.

Value of pH at every point is calculated by using the relation between pH and pOH.

pH=14pOH

The relation between Ka and Kb for weak acid and its conjugate base is given as,

Kw=(Ka)(Kb) (3)

Explanation of Solution

The pOH value before the titration can be calculated by using Kb and its relation with H3O+ ion concentration.

Kb=[OH](eq)[BH+](eq)[B](eq)

Given:

Refer to the Apendix I in the textbook for the value of Kb.

The value of Kb for ethylamine is 4.3×104.

The value of Kw for water is 1.0×1014.

The pKb value is calculated as follows;

pKb=log(Kb)

Substitute, 4.3×104 for Kb.

pKa=log(4.3×104)=3.36

Therefore, pKb value of ethylamine is 3.36.

The initial concentration of C2H5NH2 is 0.150 M.

The initial concentration of HCl is 0.100 M.

Volume of the solvent is 50 mL.

Therefore volume of the solvent is 0.050 L.

ICE table (1) gives the dissociation of C2H5NH2.

EquationC2H5NH2(aq)+H2O(l)OH(aq)+C2H5NH3+(aq)Initial(molL1)0.15000Change(molL1)x+x+xAfterreaction(molL1)0.150x+x+x

From ICE table (1),

Concentration of C2H5NH2 left after reaction is (0.150x)molL1.

Concentration of C2H5NH3+ produced after the reaction is xmolL1.

Concentration of OH produced after the reaction is xmolL1.

There is an approximation, that the value of x is very small as comparison to 0.150 thus it can be neglected with respect to it.

Therefore, Concentration of C2H5NH2 left after reaction is 0.150 molL1.

The hydroxide ion concentration is calculated by expression,

Kb=[OH](eq)[BH+](eq)[B](eq)

Substitute x for [OH](eq), x for [BH+](eq), 0.150 for [B](eq) and 4.3×104 for Ka.

4.3×104=(x)(x)(0.150)x2=(4.3×104)(0.150)

Rearrange for x,

x=0.80×102

Therefore, concentration of hydroxide ion is 0.00803 molL1.

Calculate the pOH value by using following expression,

pOH = log[OH]

Substitute 0.00803 for [OH].

pOH = log(0.00803)=2.1

The value of pOH for the original solution of C2H5NH2 is 2.1.

pH = 142.1=11.9

Therefore value of pH for the original solution is 11.9.

 (b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The value of pH at midpoint of the titration has to be calculated.

Concept introduction:

Titration is a quantitative method to determine the quantity of an acid or base in a solution. This method is used to determine the concentration an acid in the solution by titrating it against a base.

For strong acid-weak base titration the pH value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of C2H5NH2 with HCl is represented as,

C2H5NH2(aq)+ H3O+(aq)H2O(l)+C2H5NH3+(aq)

Calculation of pH at various points is done as follows,

(1) The pOH value before the titration can be calculated by using Kb and its relation with OH ion concentration.

Kb=[OH](eq)[BH+](eq)[B](eq) (1)

(2) The pH calculation just before the equivalence point,

As the addition of HCl is done there will be formation of buffer solution C2H5NH2/C2H5NH3+. The pOH calculation for buffer solution is done by using Henderson-Hasselbalch equation.

pOH=pKb+log[conjugate acid][base] (2)

At the midpoint of the titration, concentration of base and its conjugate acid is equal. Therefore pOH value at midpoint will be given as

pOH=pKb+log[conjugate acid][base]

Substitute, [conjugateacid]for[base].

pOH=pKb+log[base][base]=pKb+log(1)=pKb+0=pKb

Therefore, pOH value at midpoint is equal to pKb.

(3) The pOH calculation the equivalence point.

At equivalence point all the acid will be neutralized, and there will be only C2H5NH3+. The H3O+ will be produced due to the hydrolysis of conjugate acid at equivalence point. The hydrolysis equilibrium is represented as,

C2H5NH3+(aq)+H2O(l)H3O+(aq)+C2H5NH2(aq)

By using the value of Ka for the ethylamine, concentration of  H3O+ can be calculated.

Value of pH at every point is calculated by using the relation between pH and pOH.

pH=14pOH

The relation between Ka and Kb for weak acid and its conjugate base is given as,

Kw=(Ka)(Kb) (3)

Explanation of Solution

The pOH value at midpoint is equal to pKb.

Given:

The pKb value of ethylamine is 3.36 calculated in part (a).

Therefore, the value of pOH at midpoint is 3.36.

The value of pH is calculated.

pH = 143.36=10.64

The value of pH at the midpoint of the titration is 10.64.

 (c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The value of pH when 75 % volume of the HCl added to ethylamine solution has to be calculated.

Concept introduction:

Titration is a quantitative method to determine the quantity of an acid or base in a solution. This method is used to determine the concentration an acid in the solution by titrating it against a base.

For strong acid-weak base titration the pH value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of C2H5NH2 with HCl is represented as,

C2H5NH2(aq)+ H3O+(aq)H2O(l)+C2H5NH3+(aq)

Calculation of pH at various points is done as follows,

(1) The pOH value before the titration can be calculated by using Kb and its relation with OH ion concentration.

Kb=[OH](eq)[BH+](eq)[B](eq) (1)

(2) The pH calculation just before the equivalence point,

As the addition of HCl is done there will be formation of buffer solution C2H5NH2/C2H5NH3+. The pOH calculation for buffer solution is done by using Henderson-Hasselbalch equation.

pOH=pKb+log[conjugate acid][base] (2)

At the midpoint of the titration, concentration of base and its conjugate acid is equal. Therefore pOH value at midpoint will be given as

pOH=pKb+log[conjugate acid][base]

Substitute, [conjugateacid]for[base].

pOH=pKb+log[base][base]=pKb+log(1)=pKb+0=pKb

Therefore, pOH value at midpoint is equal to pKb.

(3) The pOH calculation the equivalence point.

At equivalence point all the acid will be neutralized, and there will be only C2H5NH3+. The H3O+ will be produced due to the hydrolysis of conjugate acid at equivalence point. The hydrolysis equilibrium is represented as,

C2H5NH3+(aq)+H2O(l)H3O+(aq)+C2H5NH2(aq)

By using the value of Ka for the ethylamine, concentration of  H3O+ can be calculated.

Value of pH at every point is calculated by using the relation between pH and pOH.

pH=14pOH

The relation between Ka and Kb for weak acid and its conjugate base is given as,

Kw=(Ka)(Kb) (3)

Explanation of Solution

The pH value when 75 % volume of the HCl added to ethylamine solution is calculated by Henderson –Hasselbalch equation.

Given:

The volume of the HCl used is 0.075 L or 75 mL to neutralize entire base.

The 75 % volume of the HCl is the 75 % of 75 mL .

(75 mL)(75)100=56.25 mL

When 56.25 mLHCl is added to ethylamine solution ICE table for the reaction between HCl and C2H5NH2 is given below,

EquationC2H5NH2(aq)+   H3O+(aq)H2O(l)+C2H5NH3+(aq)Initial(mol)0.00750.00560Change(mol)0.00560.00560.0056Afterreaction(mol)0.001900.0056

Total volume of the solution is 0.112.5 L.

The value of pOH is calculated

pOH=pKb+log[conjugate acid][base]=3.36+log(0.0056)(0.0019)=3.36+0.469=3.83

Therefore, value of pOH is 3.83.

The value of pH is calculated as

pH = 14pOH= 143.83=10.37

Therefore value of pH is 10.37 when 75 % volume of the HCl added to ethylamine solution.

(d)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The value of pH at equivalence point has to be calculated.

Concept introduction:

Titration is a quantitative method to determine the quantity of an acid or base in a solution. This method is used to determine the concentration an acid in the solution by titrating it against a base.

For strong acid-weak base titration the pH value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of C2H5NH2 with HCl is represented as,

C2H5NH2(aq)+ H3O+(aq)H2O(l)+C2H5NH3+(aq)

Calculation of pH at various points is done as follows,

(1) The pOH value before the titration can be calculated by using Kb and its relation with OH ion concentration.

Kb=[OH](eq)[BH+](eq)[B](eq) (1)

(2) The pH calculation just before the equivalence point,

As the addition of HCl is done there will be formation of buffer solution C2H5NH2/C2H5NH3+. The pOH calculation for buffer solution is done by using Henderson-Hasselbalch equation.

pOH=pKb+log[conjugate acid][base] (2)

At the midpoint of the titration, concentration of base and its conjugate acid is equal. Therefore pOH value at midpoint will be given as

pOH=pKb+log[conjugate acid][base]

Substitute, [conjugateacid]for[base].

pOH=pKb+log[base][base]=pKb+log(1)=pKb+0=pKb

Therefore, pOH value at midpoint is equal to pKb.

(3) The pOH calculation the equivalence point.

At equivalence point all the acid will be neutralized, and there will be only C2H5NH3+. The H3O+ will be produced due to the hydrolysis of conjugate acid at equivalence point. The hydrolysis equilibrium is represented as,

C2H5NH3+(aq)+H2O(l)H3O+(aq)+C2H5NH2(aq)

By using the value of Ka for the ethylamine, concentration of  H3O+ can be calculated.

Value of pH at every point is calculated by using the relation between pH and pOH.

pH=14pOH

The relation between Ka and Kb for weak acid and its conjugate base is given as,

Kw=(Ka)(Kb) (3)

Explanation of Solution

Given:

The volume of HCl used to neutralize the entire base C2H5NH2 is calculated.

Expression used for the neutralization is as follows,

C1V1=C2V2

Here,

  • C1 is the concentration of C2H5NH2
  • V1 is the volume of C2H5NH2.
  • C2 is the concentration of HCl used.
  • V2 is the volume of HCl used for the neutralization.

Substitute 0.150 molL1 for C1, 0.050 L for V1,0.100 molL1 for C2.

(0.150 molL1)(0.050L)=(0.100 molL1)V2

Rearrange for V2,

V2=(0.150 molL1)(0.050L)(0.100 molL1)=0.075 L

Therefore, volume of the HCl used is 0.075 L or 75 mL.

The calculation of moles is done by using the expression,

Numberof moles=concentration(molL1)volume(L)

The ICE table (2) for the reaction between HCl and C2H5NH2 is given below,

EquationC2H5NH2(aq)+   H3O+(aq)H2O(l)+C2H5NH3+(aq)Initial(mol)0.00750.00750Change(mol)0.00750.00750.0075Afterreaction(mol)000.0075

From ICE table (2),

Number of moles of ethylamine left after reaction are 0mol.

Number of moles of conjugate acid C2H5NH3+ produced after the reaction are 0.0075 mol.

The total volume after the reaction is calculated as,

totalvolume=volumeofC2H5NH2(L) + volume of HCl(L)

totalvolume = 0.050(L)+0.075 L=0.125 L

Therefore, total volume after reaction is 0.125 L.

Concentration calculations is done by using the expression,

concentration = Numberof molestotal volume(molL1) (4)

Calculate the concentration of C2H5NH3+ after reaction.

Substitute, 0.0075 mol for Numberof moles and 0.125 L for volume in equation (4).

concentration = 0.00750.125(molL1)=0.06molL1

The concentration of C2H5NH3+ after reaction is 0.06molL1.

The C2H5NH3+ produced will undergo hydrolysis in presence of water and the reaction equilibrium is written as,

C2H5NH3+(aq)+H2O(l)H3O+(aq)+C2H5NH2(aq)

The hydrolysis equilibrium is represented in ICE table (3).

EquationC2H5NH3+(aq)+H2O(l)H3O+(aq)+C2H5NH2(aq)Initial(molL1)0.0600Change(molL1)x+x+xAfterreaction(molL1)0.06x+x+x

From ICE table (3),

Concentration of C2H5NH3+ left after reaction is (0.06x)molL1.

Approximation, x is very small as compared to 0.06. Therefore it can be neglected.

So, Concentration of C2H5NH3+ left after reaction is (0.06)molL1

Concentration of ethylamine produced after the reaction is xmolL1.

Concentration of H3O+ produced after the reaction is xmolL1.

Calculate the concentration of OH by using the equation (3).

Kw=(Ka)(Kb)

Rearrange it for Ka

Ka=KwKb (5)

Substitute 4.3×104 for Kb and 1.0×1014 for Kw

Ka=1.0×10144.3×104=0.23×1010

Therefore, Ka value for ethylamine is 0.23×1010.

The expression of Ka for ethylamine from the ICE table (3) will be written as,

Ka=[C2H5NH2](eq)[H3O+](eq)[C2H5NH3+](eq) (6)

Substitute, 0.23×1010 for Ka, x for [H3O+](eq), x for [C2H5NH2](eq), (0.06) for [C2H5NH3+](eq).

0.23×1010=(x)(x)(0.06)

Rearrange for x,

x=(0.23×1010)(0.06)=0.1178×105

Therefore value of hydronium ion concentration is 0.1178×105 molL1.

Calculate the value of pH by using the expression,

pH = log[H3O+]

Substitute, 0.1178×105  for [H3O+].

pH = log(0.1178×105 )=5.928

Therefore, the value of pH at equivalence point is 5.928.

(e)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The value of pH when 10 mL HCl is added after equivalence point has to be calculated.

Concept introduction:

Titration is a quantitative method to determine the quantity of an acid or base in a solution. This method is used to determine the concentration an acid in the solution by titrating it against a base.

For strong acid-weak base titration the pH value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of C2H5NH2 with HCl is represented as,

C2H5NH2(aq)+ H3O+(aq)H2O(l)+C2H5NH3+(aq)

Calculation of pH at various points is done as follows,

(1) The pOH value before the titration can be calculated by using Kb and its relation with OH ion concentration.

Kb=[OH](eq)[BH+](eq)[B](eq) (1)

(2) The pH calculation just before the equivalence point,

As the addition of HCl is done there will be formation of buffer solution C2H5NH2/C2H5NH3+. The pOH calculation for buffer solution is done by using Henderson-Hasselbalch equation.

pOH=pKb+log[conjugate acid][base] (2)

At the midpoint of the titration, concentration of base and its conjugate acid is equal. Therefore pOH value at midpoint will be given as

pOH=pKb+log[conjugate acid][base]

Substitute, [conjugateacid]for[base].

pOH=pKb+log[base][base]=pKb+log(1)=pKb+0=pKb

Therefore, pOH value at midpoint is equal to pKb.

(3) The pOH calculation the equivalence point.

At equivalence point all the acid will be neutralized, and there will be only C2H5NH3+. The H3O+ will be produced due to the hydrolysis of conjugate acid at equivalence point. The hydrolysis equilibrium is represented as,

C2H5NH3+(aq)+H2O(l)H3O+(aq)+C2H5NH2(aq)

By using the value of Ka for the ethylamine, concentration of  H3O+ can be calculated.

Value of pH at every point is calculated by using the relation between pH and pOH.

pH=14pOH

The relation between Ka and Kb for weak acid and its conjugate base is given as,

Kw=(Ka)(Kb) (3)

Explanation of Solution

After equivalence point the value of pH is calculated by using Henderson-Hasselbalch equation.

Given:

When 10 mL extra solution of HCl is used after equivalence point.

The total volume of HCl added is 75+10=85 mL or 0.085 L.

The ICE table for the reaction between HCl and C2H5NH2 is given below,

EquationC2H5NH2(aq)+H3O+(aq)H2O(l)+C2H5NH3+(aq)Initial(mol)0.00750.008500Change(mol)0.00750.00750+0.0075Afterreaction(mol)00.001000.0075

After titration volume will be equal for both acid and base. The total volume will be 0.135 L.

After the equivalence point, the excess concentration H3O+ will present at the equilibrium. The hydrolysis of C2H5NH3+ will produce a very small amount of H3O+ ions, that amount can be neglected.

Concentration of hydronium ion after equivalence point is calculated.

concentration=0.0010 mol0.135 L=0.00740 molL1

Therefore, concentration of hydronium ions after reaction is 0.00740 molL1.

Calculate the pH

pH=log[H3O+]=log(0.00740 )=2.13

Therefore value of pH is 2.13 after addition of 10 mLHCl solution after equivalence point.

 (f)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

 Titration curve has to be plotted.

Concept introduction:

Titration is a quantitative method to determine the quantity of an acid or base in a solution. This method is used to determine the concentration an acid in the solution by titrating it against a base.

For strong acid-weak base titration the pH value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of C2H5NH2 with HCl is represented as,

C2H5NH2(aq)+ H3O+(aq)H2O(l)+C2H5NH3+(aq)

Calculation of pH at various points is done as follows,

(1) The pOH value before the titration can be calculated by using Kb and its relation with OH ion concentration.

Kb=[OH](eq)[BH+](eq)[B](eq) (1)

(2) The pH calculation just before the equivalence point,

As the addition of HCl is done there will be formation of buffer solution C2H5NH2/C2H5NH3+. The pOH calculation for buffer solution is done by using Henderson-Hasselbalch equation.

pOH=pKb+log[conjugate acid][base] (2)

At the midpoint of the titration, concentration of base and its conjugate acid is equal. Therefore pOH value at midpoint will be given as

pOH=pKb+log[conjugate acid][base]

Substitute, [conjugateacid]for[base].

pOH=pKb+log[base][base]=pKb+log(1)=pKb+0=pKb

Therefore, pOH value at midpoint is equal to pKb.

(3) The pOH calculation the equivalence point.

At equivalence point all the acid will be neutralized, and there will be only C2H5NH3+. The H3O+ will be produced due to the hydrolysis of conjugate acid at equivalence point. The hydrolysis equilibrium is represented as,

C2H5NH3+(aq)+H2O(l)H3O+(aq)+C2H5NH2(aq)

By using the value of Ka for the ethylamine, concentration of  H3O+ can be calculated.

Value of pH at every point is calculated by using the relation between pH and pOH.

pH=14pOH

The relation between Ka and Kb for weak acid and its conjugate base is given as,

Kw=(Ka)(Kb) (3)

Explanation of Solution

The titration curve is drawn by using the pH value at different points.  The graph parameters are x axis the volume of HCl and pH value on y axis.

Table gives the value of pH at different volume addition of HCl and corresponding point on the curve.

Points volume of HCl          (mL)pHa011.9b37.510.64c56.2510.37d755.93e852.13

Chemistry & Chemical Reactivity, Chapter 17, Problem 107IL

Graph between pH and volume of HCl added.

(g)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

A best indicator that can be used to detect the equivalence point has to be chosen.

Concept introduction:

Titration is a quantitative method to determine the quantity of an acid or base in a solution. This method is used to determine the concentration an acid in the solution by titrating it against a base.

For strong acid-weak base titration the pH value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of C2H5NH2 with HCl is represented as,

C2H5NH2(aq)+ H3O+(aq)H2O(l)+C2H5NH3+(aq)

Calculation of pH at various points is done as follows,

(1) The pOH value before the titration can be calculated by using Kb and its relation with OH ion concentration.

Kb=[OH](eq)[BH+](eq)[B](eq) (1)

(2) The pH calculation just before the equivalence point,

As the addition of HCl is done there will be formation of buffer solution C2H5NH2/C2H5NH3+. The pOH calculation for buffer solution is done by using Henderson-Hasselbalch equation.

pOH=pKb+log[conjugate acid][base] (2)

At the midpoint of the titration, when concentration of base and its conjugate acid is equal. Therefore pOH value at midpoint will be given as

pOH=pKb+log[conjugate acid][base]

Substitute, [conjugateacid]for[base].

pOH=pKb+log[base][base]=pKb+log(1)=pKb+0=pKb

Therefore, pOH value at midpoint is equal to pKb.

(3) The pOH calculation the equivalence point.

At equivalence point all the acid will be neutralized, and there will be only C2H5NH3+. The H3O+ will be produced due to the hydrolysis of conjugate acid at equivalence point. The hydrolysis equilibrium is represented as,

C2H5NH3+(aq)+H2O(l)H3O+(aq)+C2H5NH2(aq)

By using the value of Ka for the ethylamine, concentration of  H3O+ can be calculated.

Value of pH at every point is calculated by using the relation between pH and pOH.

pH=14pOH

The relation between Ka and Kb for weak acid and its conjugate base is given as,

Kw=(Ka)(Kb) (3)

Explanation of Solution

Alizarin or Bromcresol purple are the best indicator for the titration between ethylamine and HCl.  As Bromcresol purple shows colour change in the pH region 5.26.6 (approx) and Alizarin shows colour change in the pH region 5.87.3 (approx). The value of pH at equivalence point is 5.928. Thus both indicators can be used to detect the equivalence point.

Alizarin or Bromcresol purple are the best indicator for the titration between ethylamine and HCl.

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Chapter 17 Solutions

Chemistry & Chemical Reactivity

Ch. 17.3 - The titration of 0.100 M acetic acid with 0.100 M...Ch. 17.3 - Calculate the pH after 75.0 mL of 0.100 M HO has...Ch. 17.3 - 1. What is the pH after 25.0 ml of 0.100 M NaOH...Ch. 17.3 - 2. What is the pH at the equivalence point in the...Ch. 17.3 - Prob. 3RCCh. 17.3 - Phosphate ions are abundant in cells, both as the...Ch. 17.3 - A typical total phosphate concentration in a cell,...Ch. 17.4 - The barium ion concentration, [Ba2+], in a...Ch. 17.4 - Calculate the solubility of AgCN in moles per...Ch. 17.4 - Calculate the solubility of Ca(OH)2 in moles per...Ch. 17.4 - Calculate the solubility of BaSO4 (a) in pure...Ch. 17.4 - 12. Calculate the solubility of Zn(CN)2 at 25°C...Ch. 17.4 - What is the Ksp expression for silver carbonate?...Ch. 17.4 - 2. Using Ksp values, predict which salt in each...Ch. 17.4 - What is the solubility of PbSO4 in water at 25C?...Ch. 17.4 - 4. What is the solubility of PbSO4 in water at...Ch. 17.4 - Prob. 5RCCh. 17.5 - Solid Pbl2 (Ksp = 9.8 109) is placed in a beaker...Ch. 17.5 - Prob. 2CYUCh. 17.5 - Prob. 3CYUCh. 17.5 - 1. Will SrSO4 precipitate from a solution...Ch. 17.6 - Silver nitrate (0.0050 mol) is added to 1.00 L of...Ch. 17.6 - 1. Iron(II) chloride (0.025 mol) is added to 1.00...Ch. 17.7 - Calculate the value of the equilibrium constant,...Ch. 17.7 - 1. What is the equilibrium constant for the...Ch. 17.7 - Prob. 1QCh. 17.7 - What is the minimum volume of 0.0071 M NaCN(aq)...Ch. 17.7 - Use the formation constant of [Au(CN)2] in...Ch. 17.7 - Silver undergoes similar reactions as those shown...Ch. 17.7 - Write a balanced chemical equation for the...Ch. 17 - Does the pH of the solution increase, decrease or...Ch. 17 - Does the pH of the solution increase, decrease, or...Ch. 17 - What is the pH of a solution that consists of 0.20...Ch. 17 - What is the pH of 0.15 M acetic acid to which 1.56...Ch. 17 - What is the pH of the solution that results from...Ch. 17 - What is the pH of the solution that results from...Ch. 17 - What is the pH of the buffer solution that...Ch. 17 - Lactic acid (CH3CHOHCO2H) is found in sour milk,...Ch. 17 - What mass of sodium acetate, NaCH3CO2, must he...Ch. 17 - What mass of ammonium chloride, NH4Cl, must be...Ch. 17 - Calculate the pH of a solution that has an acetic...Ch. 17 - Calculate the pH of a solution that has an...Ch. 17 - What must the ratio of acetic acid to acetate ion...Ch. 17 - What must the ratio of H2PO4 to HPO42 be to have a...Ch. 17 - A buffer is composed of formic acid and its...Ch. 17 - A buffer solution is composed of 1.360 g of KH2PO4...Ch. 17 - Which of the following combinations would be the...Ch. 17 - Which of the following combinations would be the...Ch. 17 - Describe how to prepare a buffer solution from...Ch. 17 - Describe how to prepare a buffer solution from NH3...Ch. 17 - Determine the volume (in mL) of 1.00 M NaOH that...Ch. 17 - Determine the volume (in mL) of 1.00 M HC1 that...Ch. 17 - A buffer solution was prepared by adding 4.95 g of...Ch. 17 - You dissolve 0.425 g of NaOH in 2.00 L of a buffer...Ch. 17 - A buffer solution is prepared by adding 0.125 mol...Ch. 17 - What is the pH change when 20.0 mL of 0.100 M NaOH...Ch. 17 - Phenol, C6H5OH, is a weak organic acid. Suppose...Ch. 17 - Assume you dissolve 0.235 g of the weak acid...Ch. 17 - You require 36.78 mL of 0.0105 M HCl to reach the...Ch. 17 - A titration of 25.0 mL of a solution of the weak...Ch. 17 - Without doing detailed calculations, sketch the...Ch. 17 - Without doing detailed calculations, sketch the...Ch. 17 - You titrate 25.0 mL of 0.10 M NH3 with 0.10 M HCl....Ch. 17 - Using Figure 17.11, suggest an indicator to use in...Ch. 17 - Using Figure 17.11, suggest an indicator to use in...Ch. 17 - Name two insoluble salts of each of the following...Ch. 17 - Prob. 38PSCh. 17 - Using the solubility guidelines (Figure 3.10),...Ch. 17 - Predict whether each of the fallowing is insoluble...Ch. 17 - For each of the following insoluble salts, (1)...Ch. 17 - Prob. 42PSCh. 17 - When 1.55 g of solid thallium(I) bromide is added...Ch. 17 - At 20 C, a saturated aqueous solution of silver...Ch. 17 - When 250 mg of SrF2, strontium fluoride, is added...Ch. 17 - Calcium hydroxide, Ca(OH)2, dissolves in water to...Ch. 17 - You add 0.979 g of Pb(OH)2 to 1.00 L of pure water...Ch. 17 - You place 1.234 g of solid Ca(OH)2 in 1.00 L of...Ch. 17 - Estimate the solubility of silver iodide in pure...Ch. 17 - What is the molar concentration of Au+(aq) in a...Ch. 17 - Prob. 51PSCh. 17 - Estimate the solubility of lead(II) bromide (a) in...Ch. 17 - The Ksp value for radium sulfate, RaSO4, is 4.2 ...Ch. 17 - If 55 mg of lead(II) sulfate is placed in 250 mL...Ch. 17 - Prob. 55PSCh. 17 - Prob. 56PSCh. 17 - Calculate the molar solubility of silver...Ch. 17 - Calculate the solubility of silver bromide, AgBr,...Ch. 17 - Compare the solubility, in milligrams per...Ch. 17 - What is the solubility, in milligrams per...Ch. 17 - Calculate the solubility, in moles per liter, of...Ch. 17 - Calculate the solubility, in moles per liter, of...Ch. 17 - Which insoluble compound in each pair should be...Ch. 17 - Which compound in each pair is more soluble in...Ch. 17 - You have a solution that has a lead(II) ion...Ch. 17 - Sodium carbonate is added to a solution in which...Ch. 17 - If the concentration of Zn2+ in 10.0 mL of water...Ch. 17 - You have 95 mL of a solution that has a lead(II)...Ch. 17 - Prob. 69PSCh. 17 - Will a precipitate of Mg(OH)2 form when 25.0 mL of...Ch. 17 - Zinc hydroxide is amphoteric (Section 16.10). Use...Ch. 17 - Solid silver iodide, AgI, can be dissolved by...Ch. 17 - What amount of ammonia (moles) must be added to...Ch. 17 - Can you dissolve 15.0 mg of AuCl in 100.0 mL of...Ch. 17 - What is the solubility of AgCl (a) in pure water...Ch. 17 - Prob. 76PSCh. 17 - Prob. 77GQCh. 17 - Prob. 78GQCh. 17 - Prob. 79GQCh. 17 - Calculate the hydronium ion concentration and the...Ch. 17 - Calculate the hydronium ion concentration and the...Ch. 17 - For each of the following cases, decide whether...Ch. 17 - Prob. 83GQCh. 17 - A sample of hard water contains about 2.0 103 M...Ch. 17 - What is the pH of a buffer solution prepared from...Ch. 17 - Prob. 86GQCh. 17 - Describe the effect on the pH of the following...Ch. 17 - What volume of 0.120 M NaOH must be added to 100....Ch. 17 - A buffer solution is prepared by dissolving 1.50 g...Ch. 17 - What volume of 0.200 M HCl must be added to 500.0...Ch. 17 - What is the equilibrium constant for the following...Ch. 17 - Calculate the equilibrium constant for the...Ch. 17 - Prob. 93GQCh. 17 - The solubility product constant for calcium...Ch. 17 - In principle, the ions Ba2+ and Ca2+ can be...Ch. 17 - A solution contains 0.10 M iodide ion, I, and 0.10...Ch. 17 - A solution contains Ca2+ and Pb2+ ions, both at a...Ch. 17 - Prob. 98GQCh. 17 - Prob. 99GQCh. 17 - Prob. 100GQCh. 17 - Each pair of ions below is found together in...Ch. 17 - Each pair of ions below is found together in...Ch. 17 - The cations Ba2+ and Sr2+ can be precipitated as...Ch. 17 - You will often work with salts of Fe3+, Pb2+, and...Ch. 17 - Aniline hydrochloride, (C6H5NH3)Cl, is a weak...Ch. 17 - The weak base ethanolamine. HOCH2CH2NH2, can be...Ch. 17 - For the titration of 50.0 mL of 0.150 M...Ch. 17 - A buffer solution with it pH of 12.00 consists of...Ch. 17 - To have a buffer with a pH of 2.50, what volume of...Ch. 17 - What mass of Na3PO4 must be added to 80.0 mL of...Ch. 17 - You have a solution that contains AgNO3, Pb(NO3)2,...Ch. 17 - Prob. 112ILCh. 17 - Suggest a method for separating a precipitate...Ch. 17 - Prob. 114SCQCh. 17 - Prob. 115SCQCh. 17 - Two acids, each approximately 0.01 M in...Ch. 17 - Composition diagrams, commonly known as alpha...Ch. 17 - The composition diagram, or alpha plot, for the...Ch. 17 - The chemical name for aspirin is acetylsalicylic...Ch. 17 - Prob. 120SCQ
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