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Chapter 15: Problem 112

Consider 1.0 \(\mathrm{L}\) of a solution that is 0.85 $\mathrm{M} \mathrm{HOC}_{6} \mathrm{H}_{5}\( and 0.80 \)\mathrm{M} \mathrm{NaOC}_{6} \mathrm{H}_{5} .\left(K_{\mathrm{a}} \text { for } \mathrm{HOC}_{6} \mathrm{H}_{5}=1.6 \times 10^{-10} .\right)$ a. Calculate the pH of this solution. b. Calculate the pH after 0.10 mole of \(\mathrm{HCl}\) has been added to the original solution. Assume no volume change on addition of HCl. c. Calculate the pH after 0.20 mole of NaOH has been added to the original buffer solution. Assume no volume change on addition of NaOH.

Short Answer

Expert verified
The initial pH of the buffer solution is \(9.77\). After adding \(0.10\) mole of HCl, the pH becomes \(9.66\). Finally, after adding \(0.20\) mole of NaOH, the pH is \(10.02\).

Step by step solution

01

Identify the initial concentrations of the buffer solution

For this step, we already know that the solution is 0.85 M HOC₆H₅ (acid) and 0.80 M NaOC₆H₅ (base). The Kₐ for HOC₆H₅ is 1.6 × 10^(-10).
02

Calculate the initial pH of the buffer solution

To calculate the initial pH, we'll use the Henderson-Hasselbalch equation: pH = pKₐ + log ([A⁻]/[HA]) First, we need to find pKₐ. pKₐ = -log(Kₐ), so pKₐ = -log(1.6 × 10^(-10)) = 9.80. Now, we can find the pH: pH = 9.80 + log(0.80/0.85) = 9.80 + (-0.03) = \(9.77\).
03

Calculate the resulting concentrations after adding HCl

We have added 0.10 mole of HCl, which will react with NaOC₆H₅ to form HOC₆H₅. We can use the following stoichiometric relation: NaOC₆H₅ + HCl → NaCl + HOC₆H₅ Moles of base consumed = 0.10 mol New moles of HOC₆H₅ = 0.85 + 0.10 = 0.95 New moles of NaOC₆H₅ = 0.80 - 0.10 = 0.70 Now calculate the new concentrations from the new moles (assuming no volume change): [HOC₆H₅] = 0.95 mol/L [NaOC₆H₅] = 0.70 mol/L
04

Calculate the pH after adding HCl

Now, we can use the Henderson-Hasselbalch equation again for part b: pH = pKₐ + log ([A⁻]/[HA]) pH = 9.80 + log(0.70/0.95) = 9.80 - 0.14 = \(9.66\).
05

Calculate the resulting concentrations after adding NaOH

We have added 0.20 mole of NaOH, which will react with HOC₆H₅ to form NaOC₆H₅. HOC₆H₅ + NaOH → NaOC₆H₅ + H2O Moles of acid consumed = 0.20 mol New moles of HOC₆H₅ = 0.85 - 0.20 = 0.65 New moles of NaOC₆H₅ = 0.80 + 0.20 = 1.00 Now calculate the new concentrations from the new moles (assuming no volume change): [HOC₆H₅] = 0.65 mol/L [NaOC₆H₅] = 1.00 mol/L
06

Calculate the pH after adding NaOH

Now, we can use the Henderson-Hasselbalch equation again for part c: pH = pKₐ + log ([A⁻]/[HA]) pH = 9.80 + log(1.00/0.65) = 9.80 + 0.22 = \(10.02\). To summarize the results: a. The initial pH of the buffer solution is 9.77. b. The pH after adding 0.10 mole of HCl is 9.66. c. The pH after adding 0.20 mole of NaOH is 10.02.

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Most popular questions from this chapter

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)\( by 0.200\)M \mathrm{HNO}_{3}\( . Calculate the \)\mathrm{pH}$ of the resulting solution after the following volumes of \(\mathrm{HNO}_{3}\) have been added. $$ \begin{array}{ll}{\text { a. } 0.0 \mathrm{mL}} & {\text { d. } 40.0 \mathrm{mL}} \\ {\text { b. } 20.0 \mathrm{mL}} & {\text { e. } 50.0 \mathrm{mL}} \\ {\text { c. } 25.0 \mathrm{mL}} & {\text { f. } 100.0 \mathrm{mL}}\end{array} $$

Mixing together solutions of acetic acid and sodium hydroxide can make a buffered solution. Explain. How does the amount of each solution added change the effectiveness of the buffer?

A certain indicator HIn has a \(\mathrm{p} K_{\mathrm{a}}\) of 3.00 and a color change becomes visible when 7.00\(\%\) of the indicator has been converted to \(\mathrm{In}^{-}\) . At what pH is this color change visible?

a. Calculate the \(\mathrm{pH}\) of a buffered solution that is 0.100 \(\mathrm{M}\) in \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\) (benzoic acid, \(K_{\mathrm{a}}=6.4 \times 10^{-5} )\) and 0.100 \(\mathrm{M}\) in \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{Na}\) b. Calculate the pH after 20.0\(\%\) (by moles) of the benzoic acid is converted to benzoate anion by addition of a strong base. Use the dissociation equilibrium $$ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}(a q) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}(a q)+\mathrm{H}^{+}(a q) $$ to calculate the pH. c. Do the same as in part b, but use the following equilibrium to calculate the \(\mathrm{pH} :\) $$ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}(a q)+\mathrm{OH}^{-}(a q) $$ d. Do your answers in parts \(b\) and \(c\) agree? Explain.

Consider a solution formed by mixing 50.0 \(\mathrm{mL}\) of 0.100 \(\mathrm{M}\) \(\mathrm{H}_{2} \mathrm{SO}_{4}, 30.0 \mathrm{mL}\) of 0.100 \(\mathrm{M}\) HOCl, 25.0 \(\mathrm{mL}\) of 0.200 \(\mathrm{M}\) \(\mathrm{NaOH}, 25.0 \mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2},\) and 10.0 \(\mathrm{mL}\) of 0.150 \(M \mathrm{KOH}\) . Calculate the \(\mathrm{pH}\) of this solution.
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