Concept explainers
(a)
Interpretation:
The pH of the given points of the titration of
Prior to the addition of any
Concept Introduction:
pOH definition:
The pOH of a solution is defined as the negative base-10 logarithm of the hydroxide ion
Relationship between pH and pOH:
(a)
Answer to Problem 16.144QP
The pH of the given points of the titration of
The pH prior to the addition of any
Explanation of Solution
To Calculate: The pH prior to the addition of any
Given data:
A 0.150 M solution of
The volume of
The concentration of
The concentration of
The
The pH prior to the addition of any
Construct an equilibrium table for the hydrolysis of hypochlorite ion as follows,
|
|||
Initial |
0.150
0.150-x |
0.00 | 0.00 |
Change |
|
|
|
Equilibrium |
x | x |
The
The
Now substitute equilibrium concentrations into the equilibrium-constant expression.
Here, x gives the concentration of hydroxide ion
Finally, calculate pOH and then the pH as follows,
The pH is calculated as follows,
The pH prior to the addition of any
(b)
Interpretation:
The pH of the given points of the titration of
Halfway to the equivalence point
Concept Introduction:
pOH definition:
The pOH of a solution is defined as the negative base-10 logarithm of the hydroxide ion
Relationship between pH and pOH:
(b)
Answer to Problem 16.144QP
The pH of the given points of the titration of
The pH halfway to the equivalence point is 7.45
Explanation of Solution
To Calculate: The pH halfway to the equivalence point
At the half-way point,
This gives,
The pOH is calculated from the obtained hydroxide ion concentration as follows,
The pH is calculated as follows,
The pH halfway to the equivalence point was calculated as 7.45
(c)
Interpretation:
The pH of the given points of the titration of
At the equivalence point
Concept Introduction:
pOH definition:
The pOH of a solution is defined as the negative base-10 logarithm of the hydroxide ion
Relationship between pH and pOH:
(c)
Answer to Problem 16.144QP
The pH of the given points of the titration of
The pH at the equivalence point is 4.34
Explanation of Solution
To Calculate: The pH at the equivalence point
Calculate the volume of
The volume of
Hence, the total volume is as follows,
The equilibrium reaction is,
The moles of
The concentration of
Substitute into the equilibrium-constant expression and solve.
Here, x gives the hydronium ion concentration.
The pH at the equivalence point was calculated as 4.34
(d)
Interpretation:
The pH of the given points of the titration of
After 5.00 mL of
Concept Introduction:
pOH definition:
The pOH of a solution is defined as the negative base-10 logarithm of the hydroxide ion
Relationship between pH and pOH:
(d)
Answer to Problem 16.144QP
The pH of the given points of the titration of
The pH after 5.00 mL of
Explanation of Solution
To Calculate: The pH after 5.00 mL of
The total volume after the addition of 5.00 mL of
Calculate the concentration of excess
The hydronium ion concentration and the pH are:
The pH is calculated as follows,
The pH after 5.00 mL of
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Chapter 16 Solutions
Lab Manual Experiments in General Chemistry
- A quantity of 0.25 M sodium hydroxide is added to a solution containing 0.15 mol of acetic acid. The final volume of the solution is 375 mL and the pH of this solution is 4.45. a What is the molar concentration of the sodium acetate? b How many milliliters of sodium hydroxide were added to the original solution? c What was the original concentration of the acetic acid?arrow_forwardSodium benzoate, NaC7H5O2, is used as a preservative in foods. Consider a 50.0-mL sample of 0.250 M NaC7H5O2 being titrated by 0.200 M HBr. Calculate the pH of the solution: a when no HBr has been added; b after the addition of 50.0 mL of the HBr solution; c at the equivalence point; d after the addition of 75.00 mL of the HBr solution. The Kb value for the benzoate ion is 1.6 1010.arrow_forwardTwo samples of 1.00 M HCl of equivalent volumes are prepared. One sample is titrated to the equivalence point with a 1.00 M solution of sodium hydroxide, while the other sample is titrated to the equivalence point with a 1.00 M solution of calcium hydroxide. a Compare the volumes of sodium hydroxide and calcium hydroxide required to reach the equivalence point for each titration. b Determine the pH of each solution halfway to the equivalence point. c Determine the pH of each solution at the equivalence point.arrow_forward
- A solution made up of 1.0 M NH3 and 0.50 M (NH4)2SO4 has a pH of 9.26. a Write the net ionic equation that represents the reaction of this solution with a strong acid. b Write the net ionic equation that represents the reaction of this solution with a strong base. c To 100. mL of this solution, 10.0 mL of 1.00 M HCl is added. How many moles of NH3 and NH4+ are present in the reaction system before and after the addition of the HCl? What is the pH of the resulting solution? d Why did the pH change only slightly upon the addition of HCl?arrow_forwardA solution of weak base is titrated to the equivalence point with a strong acid. Which one of the following statements is most likely to be correct? a The pH of the solution at the equivalence point is 7.0. b The pH of the solution is greater than 13.0. c The pH of the solution is less than 2.0. d The pH of the solution is between 2.0 and 7.0. e The pH of the solution is between 7.0 and 13.0. The reason that best supports my choosing the answer above is a Whenever a solution is titrated with a strong acid, the solution will be very acidic. b Because the solution contains a weak base and the acid (titrant) is used up at the equivalence point, the solution will be basic. c Because the solution contains the conjugate acid of the weak base at the equivalence point, the solution will be acidic.arrow_forwardConsider all acid-base indicators discussed in this chapter. Which of these indicators would be suitable for the titration of each of these? (a) NaOH with HClO4 (b) acetic acid with KOH (c) NH3 solution with HBr (d) KOH with HNO3 Explain your choices.arrow_forward
- How many grams of HI should be added to 265 mL of 0.215 M HCI so that the resulting solution has a pH of 0.38? Assume that the addition of HI does not change the volume of the resulting solution.arrow_forwardEstimate the pH that results when the following two solutions are mixed. a) 50 mL of 0.3 M CH3COOH and 50 mL of 0.4 M KOH b) 100 mL of 0.3 M CH3COOH and 50 mL of 0.4 M NaOH c) 150 mL of 0.3 M CH3COOH and 100 mL of 0.3 M Ba(OH)2 d) 200 mL of 0.3 M CH3COOH and 100 mL of 0.3 M Ba(OH)2arrow_forwardCalculate the pH of solutions that are 0.25 M formic acid and 0.40 M sodium formate. 0.50 M benzoic acid and 0.15 M sodium benzoate.arrow_forward
- The pH of Mixtures of Acid, Base, and Salt Solutions a When 0.10 mol of the ionic solid NaX, where X is an unknown anion, is dissolved in enough water to make 1.0 L of solution, the pH of the solution is 9.12. When 0.10 mol of the ionic solid ACl, where A is an unknown cation, is dissolved in enough water to make 1.0 L of solution, the pH of the solution is 7.00. What would be the pH of 1.0 L of solution that contained 0.10 mol of AX? Be sure to document how you arrived at your answer. b In the AX solution prepared above, is there any OH present? If so, compare the [OH] in the solution to the [H3O+]. c From the information presented in part a, calculate Kb for the X(aq) anion and Ka for the conjugate acid of X(aq). d To 1.0 L of solution that contains 0.10 mol of AX, you add 0.025 mol of HCl. How will the pH of this solution compare to that of the solution that contained only NaX? Use chemical reactions as part of your explanation; you do not need to solve for a numerical answer. e Another 1.0 L sample of solution is prepared by mixing 0.10 mol of AX and 0.10 mol of HCl. The pH of the resulting solution is found to be 3.12. Explain why the pH of this solution is 3.12. f Finally, consider a different 1.0-L sample of solution that contains 0.10 mol of AX and 0.1 mol of NaOH. The pH of this solution is found to be 13.00. Explain why the pH of this solution is 13.00. g Some students mistakenly think that a solution that contains 0.10 mol of AX and 0.10 mol of HCl should have a pH of 1.00. Can you come up with a reason why students have this misconception? Write an approach that you would use to help these students understand what they are doing wrong.arrow_forwardSulfanilic acid (NH2C6H4SO3H) is used in manufacturing dyes. It ionizes in water according to the equilibrium equation NH2C6H4SO3H(aq)+H2O(l)NH2C6H4SO3(aq)+H3O+(aq)Ka=5.9104 A buffer is prepared by dissolving 0.20 mol of sulfanilicacid and 0.13 mol of sodium sulfanilate (NaNH2C6H4SO3) in water and diluting to 1.00 L. Compute the pH of the solution. Suppose 0.040 mol of HCl is added to the buffer.Calculate the pH of the solution that results.arrow_forwardCalculate the mass in grams of ammonium chloride, NH4C1, that would have to be added to 500. mL of 0.10-M NH3 solution to have a pH of 9.00.arrow_forward
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