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Chapter 15: Problem 65

Calculate the \(\mathrm{pH}\) at the halfway point and at the equivalence point for each of the following titrations. a. \(100.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\left(K_{\mathrm{a}}=6.4 \times 10^{-5}\right)\) titrated by \(0.10 \mathrm{M} \mathrm{NaOH}\) b. \(100.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\left(K_{\mathrm{b}}=5.6 \times 10^{-4}\right)\) titrated by \(0.20 \mathrm{M} \mathrm{HNO}_{3}\) c. \(100.0 \mathrm{~mL}\) of \(0.50 \mathrm{M} \mathrm{HCl}\) titrated by \(0.25 \mathrm{M} \mathrm{NaOH}\)

Short Answer

Expert verified
For the titrations in this problem: a. At the halfway point, the pH is equal to 4.19, and at the equivalence point, the pH is found by first calculating the pOH using the ionization equation and then finding the pH using \(pH=14-pOH\). b. At the halfway point, find the pOH using the \(pK_{b}\), then calculate the pH using \(pH=14-pOH\). At the equivalence point, find the pOH using the concentration of the acid and \(K_a\), then calculate the pH using \(pH=14-pOH\). c. At the halfway point, find the pH using the initial concentrations of the strong acid and base. At the equivalence point, the pH is 7 since all the strong acid is neutralized by the strong base.

Step by step solution

01

At the halfway point of the titration, we have added exactly half of the required base to neutralize the weak acid. The ratio of the acid to its conjugate base will be equal to 1 and we can use the Henderson-Hasselbalch equation to calculate the pH: \(\mathrm{pH}=pK_{a}+\log\frac{[\mathrm{A}^-]}{[\mathrm{HA}]}\). In this case, since the ratio of \([\mathrm{A}^-]\) to \([\mathrm{HA}]\) is 1, the pH is equal to the \(pK_a\) of our acid. #Step 2: Calculate pKa#

To find the pH at the halfway point, we need to find the \(pK_a\) value of the acid first. We can find the \(pK_a\) by taking the negative logarithm of the given \(K_a\): \(pK_{a}=-\log(K_a)=-\log(6.4 \times 10^{-5})\). #Step 3: Calculate pH at halfway point#
02

Plugging the given values into the equation for pH at the halfway point, we get: \(pH=4.19\) #Step 4: Equivalence point#

At the equivalence point, all of the weak acid is neutralized, and we have only the conjugate base in solution. To find the pH, we will use the hydrolysis reaction of the conjugate base, \(\mathrm{A}^{-}\), with water and solve for the concentration of hydroxide ion. #Step 5: Calculate pOH#
03

Write an ionization equation involving the conjugate base: \(\mathrm{A}^- + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{HA} + \mathrm{OH}^-\). By using the fact that \(K_w = K_a K_b\), solving for the concentration of hydroxide ion, and using the equivalence point volume, we can find the pOH. #Step 6: Calculate pH at equivalence point#

Using the calculated pOH, we can find the pH at the equivalence point by subtracting the calculated pOH from 14: \(pH = 14 - pOH\). Following the same procedure for the other titrations, we have: b. \(100.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{5}\mathrm{NH}_{2}\left(K_{\mathrm{b}}=5.6 \times 10^{-4}\right)\) titrated by \(0.20 \mathrm{M} \mathrm{HNO}_{3}\)
04

Halfway point (pOH)#

At the halfway point: \(\mathrm{pOH}=pK_{b}=-\log(K_{b})=-\log(5.6 \times 10^{-4})\)
05

Halfway point (pH = 14 - pOH)#

Calculate the pH using the relationship pH and pOH: \(pH = 14 - pOH\)
06

Equivalence point (pOH)#

At the equivalence point, find the pOH using \(K_a\) and the concentration of the acid.
07

Equivalence point (pH = 14 - pOH)#

Calculate the pH using the relationship pH and pOH: \(pH = 14 - pOH\) c. \(100.0 \mathrm{~mL}\) of \(0.50 \mathrm{M} \mathrm{HCl}\) titrated by \(0.25\mathrm{M} \mathrm{NaOH}\)
08

Halfway point (pH)#

At the halfway point, find the pH by using the initial concentrations of the strong acid and base involved in the titration.
09

Equivalence point (pH)#

At the equivalence point, all the strong acid is neutralized by the strong base, forming a neutral solution with a pH of 7.

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Most popular questions from this chapter

Could a buffered solution be made by mixing aqueous solutions of \(\mathrm{HCl}\) and \(\mathrm{NaOH}\) ? Explain. Why isn't a mixture of a strong acid and its conjugate base considered a buffered solution?

A \(0.400-M\) solution of ammonia was titrated with hydrochloric acid to the equivalence point, where the total volume was 1.50 times the original volume. At what pH does the equivalence point occur?

Calculate the \(\mathrm{pH}\) after \(0.010\) mole of gaseous \(\mathrm{HCl}\) is added to \(250.0 \mathrm{~mL}\) of each of the following buffered solutions. a. \(0.050 \mathrm{M} \mathrm{NH}_{3} / 0.15 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) b. \(0.50 \mathrm{M} \mathrm{NH}_{3} / 1.50 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) Do the two original buffered solutions differ in their \(\mathrm{pH}\) or their capacity? What advantage is there in having a buffer with a greater capacity?

Consider the titration of \(100.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{H}_{2} \mathrm{NNH}_{2}\) \(\left(K_{\mathrm{b}}=3.0 \times 10^{-6}\right)\) by \(0.200 \mathrm{M} \mathrm{HNO}_{3} .\) Calculate the \(\mathrm{pH}\) of the resulting solution after the following volumes of \(\mathrm{HNO}_{3}\) have been added. a. \(0.0 \mathrm{~mL}\) b. \(20.0 \mathrm{~mL}\) c. \(25.0 \mathrm{~mL}\) d. \(40.0 \mathrm{~mL}\) e. \(50.0 \mathrm{~mL}\) f. \(100.0 \mathrm{~mL}\)

Consider the following two acids: $$ \mathrm{p} K_{\mathrm{a}_{1}}=2.98 ; \mathrm{p} K_{\mathrm{a}_{2}}=13.40 $$ $$ \begin{array}{l} \mathrm{HO}_{2} \mathrm{CCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H} \\ \text { Adipic acid } \quad \mathrm{p} K_{\mathrm{a}_{1}}=4.41 ; \mathrm{p} K_{\mathrm{a}_{2}}=5.28 \end{array} $$ In two separate experiments the pH was measured during the titration of \(5.00\) mmol of each acid with \(0.200 M \mathrm{NaOH}\). Each experiment showed only one stoichiometric point when the data were plotted. In one experiment the stoichiometric point was at \(25.00 \mathrm{~mL}\) added \(\mathrm{NaOH}\), and in the other experiment the stoichiometric point was at \(50.00 \mathrm{~mL} \mathrm{NaOH}\). Explain these results. (See Exercise \(113 .\) )
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