Chemistry: The Molecular Nature of Matter and Change - Standalone book
Chemistry: The Molecular Nature of Matter and Change - Standalone book
7th Edition
ISBN: 9780073511177
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
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Chapter 19, Problem 19.139P

(a)

Interpretation Introduction

Interpretation:

pH Of HF solution has to be calculated before NaOH is added.

Concept introduction:

Titration:

Titration is a quantitative chemical analysis to determine the concentration of an identified analyte. The titrant is the reagent which is prepared as a standard solution of known concentration volume. The titrant reacts with the analyte to determine the analyte’s concentration. The volume of the titrant reacting with analyte is called the titration volume.

Equivalence point:

Equivalence point in the titration reaction is the point where the amount of titrant added is absolutely enough to neutralize completely the analyte. The moles of titrant and the moles of analyte are same at this point.

pH:

pH is a scale used to specify the acidity or basicity a solution . It ranges from 014. pH 7.0 is considered as neutral solution, pH more than 7.0 is taken as basic solution whereas pH less than 7.0 is considered as acidic solution (at 25οC). It is the measurement of activity of free H+ and OH- in solution.

(a)

Expert Solution
Check Mark

Answer to Problem 19.139P

pH of solution before addition of titrant NaOH is 1.88.

Explanation of Solution

HF being a weak acid dissociates slowly in the solution producing less number of H+.

  HFH++F-

Let the dissociation constant be x.

Given that initially the concentration of HF is 0.25 M.

  HF  H++F-initially0.2500change-x+x+xequilibrium(0.25-x)x x

Acid dissociation constant (Ka)=[H+]×[F-][HF]

Now, substituting the values,

  6.8×10-4=x×x(0.25-x)

  (Ka=6.8×104, taken from text book)

x<<1 Hence (0.25-x)0.25

  x2=6.8×10-4×0.25

  x=0.0130M

Hence, concentration of H+,

  [H+]=0.0130M

Thus pH can be calculated as,

  pH=log[H+]

  pH=log0.0130=1.88

Hence pH of solution before addition of titrant NaOH is 1.88.

(b)

Interpretation Introduction

Interpretation:

Amount of titrant (NaOH) is required to achieve the equivalence point has to be calculated in milliliters.

Concept introduction:

Titration:

Titration is a quantitative chemical analysis to determine the concentration of an identified analyte. The titrant is the reagent which is prepared as a standard solution of known concentration volume. The titrant reacts with the analyte to determine the analyte’s concentration. The volume of the titrant reacting with analyte is called the titration volume.

Equivalence point:

Equivalence point in the titration reaction is the point where the amount of titrant added is absolutely enough to neutralize completely the analyte. The moles of titrant and the moles of analyte are same at this point.

pH:

pH is a scale used to specify the acidity or basicity a solution . It ranges from 014. pH 7.0 is considered as neutral solution, pH more than 7.0 is taken as basic solution whereas pH less than 7.0 is considered as acidic solution (at 25οC). It is the measurement of activity of free H+ and OH- in solution.

(b)

Expert Solution
Check Mark

Answer to Problem 19.139P

The amount of titrant required to achieve equivalence point is 57.115mL.

Explanation of Solution

Volume of HF taken is 35mL.

Strength of HF solution is 0.25M.

Strength of NaOH solution take is 0.1532M.

According to volumetric titration,

  [(Volumeofacidtaken)×(Strengthofacidtaken)] = [(Volumeofbasetaken)×(Strengthofbasetaken)]

Therefore,

  Volume of NaOH taken = (35mL×0.25M)(0.1532M)                                     = 57.115mL

Hence, the amount of titrant required to achieve equivalence point is 57.115mL.

(c)

Interpretation Introduction

Interpretation:

The value of pH at 0.50mL before equivalence point has to be calculated.

Concept introduction:

Titration:

Titration is a quantitative chemical analysis to determine the concentration of an identified analyte. The titrant is the reagent which is prepared as a standard solution of known concentration volume. The titrant reacts with the analyte to determine the analyte’s concentration. The volume of the titrant reacting with analyte is called the titration volume.

Equivalence point:

Equivalence point in the titration reaction is the point where the amount of titrant added is absolutely enough to neutralize completely the analyte. The moles of titrant and the moles of analyte are same at this point.

pH:

pH is a scale used to specify the acidity or basicity a solution . It ranges from 014. pH 7.0 is considered as neutral solution, pH more than 7.0 is taken as basic solution whereas pH less than 7.0 is considered as acidic solution (at 25οC). It is the measurement of activity of free H+ and OH- in solution.

(c)

Expert Solution
Check Mark

Answer to Problem 19.139P

pH at 0.50mL before equivalence point is 5.203.

Explanation of Solution

HFH++F-

Given that 35mLof0.25M solution of HF is taken.

Total no. of moles of HF taken=(35mL×0.25mol)1000mL=0.00875mole

Now according to question pH at 0.50ml before equivalence point has to be calculated.

  Hence the volume of NaOH to be added =(57.115 – 0.50 )mL. =56.615 mL

  Hence no. of moles of NaOH added(56.615ml×0.1532mol)1000ml  =0.00867mole

Number of moles of F- formed are equals to the no. of moles of NaOH added.

Hence number of moles of F- formed =0.00867moles.

  Hence excess amount of HF formed=(0.00875 - 0.00867)moles=0.00008moles.

  Volume of total solution=(35+57.115-0.50)mL     =91.615mL= 91.615×103 L

Thus the concentration of HF is calculated as,

  [HF]=(0.00008mol)91.615×103 L=0.00087M (Excess HF)

Thus the concentration of F- is calculated as,

  [F-]=(0.00867mol)91.615×103 L=0.0946M

Ka=6.8×104

Hence, pKa can be calculated as,

  pKa=logKa=log(6.8×104)=3.167

  pKa=3.167

From Henderson-Hasselbalch equation,

  pH=pKa+ log[F-][HF]

Hence, pH can be calculated as,

  pH=3.167+ log0.09460.00087

  pH=5.203

Hence, pH at 0.50mL before equivalence point is 5.203.

(d)

Interpretation Introduction

Interpretation:

pH At equivalence point has to be calculated.

Concept introduction:

Titration:

Titration is a quantitative chemical analysis to determine the concentration of an identified analyte. The titrant is the reagent which is prepared as a standard solution of known concentration volume. The titrant reacts with the analyte to determine the analyte’s concentration. The volume of the titrant reacting with analyte is called the titration volume.

Equivalence point:

Equivalence point in the titration reaction is the point where the amount of titrant added is absolutely enough to neutralize completely the analyte. The moles of titrant and the moles of analyte are same at this point.

pH:

pH is a scale used to specify the acidity or basicity a solution . It ranges from 014. pH 7.0 is considered as neutral solution, pH more than 7.0 is taken as basic solution whereas pH less than 7.0 is considered as acidic solution (at 25οC). It is the measurement of activity of free H+ and OH- in solution.

(d)

Expert Solution
Check Mark

Answer to Problem 19.139P

pH at equivalence point is 8.07.

Explanation of Solution

At equivalence point we are considering the equation:

  F-+H2OHF+OH-

At equivalence point the number of moles of  F- formed is 0.00875M.because at equivalence point the pH will be only due to the excess OH- in the medium.

  At equivalence point the total volume is=(35 + 57.115)ml =92.115ml

  Molarity ofF-(0.00875mol×1000mL)92.115mL =0.09499M

Now, Ka×Kb = Kw

Thus the acid dissociation constant of the base can be calculated as,

  Kb=1×10-146.8×10-4 (at 25oCKw=1×10-14)=1.47×10-11

  Kb=[HF]×[OH-][F-]

Say concentration of OH- at equilibrium is x.

  F-+H2OHF+OH-initially0.0949900change-x+x+xequilibrium(0.09499-x)x    x

  Kb=x×x(0.09499-x)

  x=1.182×106M

Thus the concentration of OH- is,

  [OH-]=1.182×10-6M

At 25οC,

  [H+][OH-]=Kw

Hence, concentration of H+ is,

  [H+]=Kw[OH-]

  [H+]=1×10-141.182×10-6=8.46×10-9M

Thus pH can be calculated as,

  pH=-log[H+]

  pH=log(8.46×109)=8.07

Hence pH at equivalence point is 8.07.

(e)

Interpretation Introduction

Interpretation:

Value of pH at 0.50mL has to be calculated after equivalence point.

Concept introduction:

Titration:

Titration is a quantitative chemical analysis to determine the concentration of an identified analyte. The titrant is the reagent which is prepared as a standard solution of known concentration volume. The titrant reacts with the analyte to determine the analyte’s concentration. The volume of the titrant reacting with analyte is called the titration volume.

Equivalence point:

Equivalence point in the titration reaction is the point where the amount of titrant added is absolutely enough to neutralize completely the analyte. The moles of titrant and the moles of analyte are same at this point.

pH:

pH is a scale used to specify the acidity or basicity a solution . It ranges from 014. pH 7.0 is considered as neutral solution, pH more than 7.0 is taken as basic solution whereas pH less than 7.0 is considered as acidic solution (at 25οC). It is the measurement of activity of free H+ and OH- in solution.

(e)

Expert Solution
Check Mark

Answer to Problem 19.139P

The value of pH at 0.50mL after equivalence point is 10.92.

Explanation of Solution

After equivalence point there will be excess NaOH and so the pH will be due to the excess OH- in the solution.

  After0.50ml of equivalence point the total volume =(35 + 57.115 + 0.50)mL =92.615mL

  Moles of excess NaOH=(0.50ml×0.1532mol)1000mL=7.66×10-5moles

Thus concentration of OH- is,

  [OH-]=(7.66×10-5mol)92.615×103L=8.271×10-4M

At 25οC,

  [H+][OH-]=Kw

Hence, concentration of H+ can be calculated as,

  [H+]=Kw[OH-]

  [H+]=1×10-148.271×10-4M    =1.209×10-11M

Thus pH becomes,

  pH=-log[H+]

 pH=log(1.209×1011)=10.92

The value of pH at 0.50mL after equivalence point is 10.92.

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Chapter 19 Solutions

Chemistry: The Molecular Nature of Matter and Change - Standalone book

Ch. 19.3 - Prob. 19.6AFPCh. 19.3 - Prob. 19.6BFPCh. 19.3 - Prob. 19.7AFPCh. 19.3 - Prob. 19.7BFPCh. 19.3 - Prob. 19.8AFPCh. 19.3 - Prob. 19.8BFPCh. 19.3 - Prob. 19.9AFPCh. 19.3 - Prob. 19.9BFPCh. 19.3 - Prob. 19.10AFPCh. 19.3 - Prob. 19.10BFPCh. 19.3 - Prob. 19.11AFPCh. 19.3 - Prob. 19.11BFPCh. 19.3 - Prob. 19.12AFPCh. 19.3 - Prob. 19.12BFPCh. 19.3 - An environmental technician collects a sample of...Ch. 19.3 - A lake that has a surface area of 10.0 acres (1...Ch. 19.4 - Cyanide ion is toxic because it forms stable...Ch. 19.4 - Prob. 19.13BFPCh. 19.4 - Prob. 19.14AFPCh. 19.4 - Calculate the solubility of PbCl2 in 0.75 M NaOH....Ch. 19 - Prob. 19.1PCh. 19 - Prob. 19.2PCh. 19 - Prob. 19.3PCh. 19 - The scenes below depict solutions of the same...Ch. 19 - Prob. 19.5PCh. 19 - Prob. 19.6PCh. 19 - Prob. 19.7PCh. 19 - Prob. 19.8PCh. 19 - Prob. 19.9PCh. 19 - Does the pH increase or decrease with each of the...Ch. 19 - What are the [H3O+] and the pH of a propanoic...Ch. 19 - What are the [H3O+] and the pH of a benzoic...Ch. 19 - Prob. 19.13PCh. 19 - Prob. 19.14PCh. 19 - Prob. 19.15PCh. 19 - Find the pH of a buffer that consists of 0.95 M...Ch. 19 - Prob. 19.17PCh. 19 - Prob. 19.18PCh. 19 - Prob. 19.19PCh. 19 - Find the pH of a buffer that consists of 0.50 M...Ch. 19 - A buffer consists of 0.22 M KHCO3 and 0.37 M...Ch. 19 - A buffer consists of 0.50 M NaH2PO4 and 0.40 M...Ch. 19 - What is the component concentration ratio,...Ch. 19 - Prob. 19.24PCh. 19 - Prob. 19.25PCh. 19 - Prob. 19.26PCh. 19 - Prob. 19.27PCh. 19 - A buffer that contains 0.40 M of a base, B, and...Ch. 19 - A buffer that contains 0.110 M HY and 0.220 M Y−...Ch. 19 - A buffer that contains 1.05 M B and 0.750 M BH+...Ch. 19 - A buffer is prepared by mixing 204 mL of 0.452 M...Ch. 19 - A buffer is prepared by mixing 50.0 mL of 0.050 M...Ch. 19 - Prob. 19.33PCh. 19 - Prob. 19.34PCh. 19 - Prob. 19.35PCh. 19 - Choose specific acid-base conjugate pairs to make...Ch. 19 - An industrial chemist studying bleaching and...Ch. 19 - Oxoanions of phosphorus are buffer components in...Ch. 19 - Prob. 19.39PCh. 19 - Prob. 19.40PCh. 19 - Prob. 19.41PCh. 19 - Prob. 19.42PCh. 19 - The scenes below depict the relative...Ch. 19 - Prob. 19.44PCh. 19 - What species are in the buffer region of a weak...Ch. 19 - Prob. 19.46PCh. 19 - Prob. 19.47PCh. 19 - Prob. 19.48PCh. 19 - Prob. 19.49PCh. 19 - Prob. 19.50PCh. 19 - Prob. 19.51PCh. 19 - Prob. 19.52PCh. 19 - Use figure 19.9 to find an indicator for these...Ch. 19 - Prob. 19.54PCh. 19 - Prob. 19.55PCh. 19 - Prob. 19.56PCh. 19 - Prob. 19.57PCh. 19 - Prob. 19.58PCh. 19 - Prob. 19.59PCh. 19 - Prob. 19.60PCh. 19 - Prob. 19.61PCh. 19 - Prob. 19.62PCh. 19 - Prob. 19.63PCh. 19 - Prob. 19.64PCh. 19 - Prob. 19.65PCh. 19 - Write the ion-product expressions for (a) silver...Ch. 19 - Prob. 19.67PCh. 19 - Prob. 19.68PCh. 19 - Prob. 19.69PCh. 19 - The solubility of silver carbonate is 0.032 M at...Ch. 19 - Prob. 19.71PCh. 19 - Prob. 19.72PCh. 19 - The solubility of calcium sulfate at 30°C is 0.209...Ch. 19 - Prob. 19.74PCh. 19 - Prob. 19.75PCh. 19 - Prob. 19.76PCh. 19 - Calculate the molar solubility of Ag2SO4 in (a)...Ch. 19 - Prob. 19.78PCh. 19 - Prob. 19.79PCh. 19 - Prob. 19.80PCh. 19 - Prob. 19.81PCh. 19 - Prob. 19.82PCh. 19 - Write equations to show whether the solubility of...Ch. 19 - Prob. 19.84PCh. 19 - Prob. 19.85PCh. 19 - Prob. 19.86PCh. 19 - Does any solid PbCl2 form when 3.5 mg of NaCl is...Ch. 19 - Prob. 19.88PCh. 19 - Prob. 19.89PCh. 19 - Prob. 19.90PCh. 19 - A 50.0-mL volume of 0.50 M Fe(NO3)3 is mixed with...Ch. 19 - Prob. 19.92PCh. 19 - Prob. 19.93PCh. 19 - Prob. 19.94PCh. 19 - Write a balanced equation for the reaction of in...Ch. 19 - Prob. 19.96PCh. 19 - Prob. 19.97PCh. 19 - Prob. 19.98PCh. 19 - What is [Ag+] when 25.0 mL each of 0.044 M AgNO3...Ch. 19 - Prob. 19.100PCh. 19 - Prob. 19.101PCh. 19 - Prob. 19.102PCh. 19 - When 0.84 g of ZnCl2 is dissolved in 245 mL of...Ch. 19 - When 2.4 g of Co(NO3)2 is dissolved in 0.350 L of...Ch. 19 - Prob. 19.105PCh. 19 - A microbiologist is preparing a medium on which to...Ch. 19 - As an FDA physiologist, you need 0.700 L of formic...Ch. 19 - Tris(hydroxymethyl)aminomethane [(HOCH2)3CNH2],...Ch. 19 - Water flowing through pipes of carbon steel must...Ch. 19 - Gout is caused by an error in metabolism that...Ch. 19 - In the process of cave formation (Section 19.3),...Ch. 19 - Phosphate systems form essential buffers in...Ch. 19 - The solubility of KCl is 3.7 M at 20°C. 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