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Chapter 15: Problem 100

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

Short Answer

Expert verified
After calculations, the initial pH of the given solution is approximately \(4.72\). After adding \(0.10\) moles of \(\mathrm{HCl}\), the pH drops to approximately \(4.44\). Finally, after adding \(0.20\) moles of \(\mathrm{NaOH}\) to the original solution, the pH increases to approximately \(5.16\).

Step by step solution

01

Write the acid dissociation equation and the expressions for \(K_a\) and \(pH\).

The acid dissociation equation for the weak acid, \(\mathrm{HOC}_6\mathrm{H}_5\), is: \[\mathrm{HOC}_6\mathrm{H}_5 + \mathrm{H_2O} \rightleftharpoons \mathrm{OC}_6\mathrm{H}_5^- + \mathrm{H_3O}^+\] Given the expression for the acid dissociation constant \(K_a\) is: \[K_a = \frac{[\mathrm{OC}_6\mathrm{H}_5^-][\mathrm{H}_3\mathrm{O}^+]}{[\mathrm{HOC}_6\mathrm{H}_5]}\] The expression for the pH of this solution can be written as: \[\mathrm{pH} = \mathrm{p}K_a + \log \frac{[\mathrm{OC}_{6}\mathrm{H}_{5}^{-}]}{[\mathrm{HOC}_{6}\mathrm{H}_{5}]}\]
02

Substitute the given values and calculate the pH.

Using the information provided: \([\mathrm{HOC}_6\mathrm{H}_5] = 0.85\,\text{M}\), \([\mathrm{OC}_6\mathrm{H}_5^-]=0.80\, \text{M}\), \(K_a = 1.6\times10^{-10}\). We can find the \(pK_a\) as follows: \[pK_a = -\log K_a = -\log (1.6 \times 10^{-10}) \] And then calculate the pH using the Henderson-Hasselbalch equation: \[\mathrm{pH} = (\mathrm{-\log (1.6 \times 10^{-10})})+ \log \frac{0.80 \,\text{M}}{0.85\, \text{M}}\] By calculating, the \(\mathrm{pH}\) of the solution is approximately \(4.72\). #b.Calculate the pH after 0.10 mole of HCl has been added to the original solution. Assume no volume change on addition of HCl.#
03

Determine the reaction between the solution and HCl and find new concentrations.

When \(\mathrm{HCl}\) is added to the solution, it reacts with the conjugate base \(\mathrm{OC}_6\mathrm{H}_5^-\). The reaction is as follows: \[\mathrm{OC}_6\mathrm{H}_5^- + \mathrm{HCl} \rightarrow \mathrm{HOC}_6\mathrm{H}_5 + \mathrm{Cl}^-\] Given that 0.10 mole of HCl is added, new concentrations are: \([\mathrm{OC}_6\mathrm{H}_5^-] = 0.80\, \text{M} - 0.10\, \text{mole/L} = 0.70\, \text{M}\) \([\mathrm{HOC}_6\mathrm{H}_5] = 0.85\, \text{M} + 0.10\, \text{mole/L} =0.95\, \text{M}\)
04

Calculate the new pH after HCl addition.

Using the new concentrations in the Henderson-Hasselbalch equation: \[\mathrm{pH} = (\mathrm{-\log (1.6 \times 10^{-10})})+ \log \frac{0.70 \,\text{M}}{0.95\, \text{M}}\] By calculating, the new \(\mathrm{pH}\) after adding \(\mathrm{HCl}\) is approximately \(4.44\). #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.#
05

Determine the reaction between the solution and NaOH and find new concentrations.

When \(\mathrm{NaOH}\) is added to the original solution, it reacts with the weak acid \(\mathrm{HOC}_6\mathrm{H}_5\). The reaction is as follows: \[\mathrm{HOC}_6\mathrm{H}_5 + \mathrm{NaOH} \rightarrow \mathrm{NaOC}_6\mathrm{H}_5 + \mathrm{H}_2\mathrm{O}\] Given that 0.20 mole of NaOH is added, new concentrations are: \([\mathrm{OC}_6\mathrm{H}_5^-] = 0.80\, \text{M} + 0.20\, \text{mole/L} = 1.00\, \text{M}\) \([\mathrm{HOC}_6\mathrm{H}_5] = 0.85\, \text{M} - 0.20\, \text{mole/L} =0.65\, \text{M}\)
06

Calculate the new pH after NaOH addition.

Using the new concentrations in the Henderson-Hasselbalch equation: \[\mathrm{pH} = (\mathrm{-\log (1.6 \times 10^{-10})})+ \log \frac{1.00 \,\text{M}}{0.65\, \text{M}}\] By calculating, the new \(\mathrm{pH}\) after adding \(\mathrm{NaOH}\) is approximately \(5.16\).

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Most popular questions from this chapter

A buffer is prepared by dissolving \(\mathrm{HONH}_{2}\) and \(\mathrm{HONH}_{3} \mathrm{NO}_{3}\) in some water. Write equations to show how this buffer neutralizes added \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\).

Consider the titration of \(150.0 \mathrm{~mL}\) of \(0.100 \mathrm{M}\) HI by \(0.250 \mathrm{M}\) \(\mathrm{NaOH} .\) a. Calculate the \(\mathrm{pH}\) after \(20.0 \mathrm{~mL}\) of \(\mathrm{NaOH}\) has been added. b. What volume of \(\mathrm{NaOH}\) must be added so that the \(\mathrm{pH}=\) \(7.00 ?\)

What volumes of \(0.50 \mathrm{M} \mathrm{HNO}_{2}\) and \(0.50 \mathrm{M} \mathrm{NaNO}_{2}\) must be mixed to prepare \(1.00 \mathrm{~L}\) of a solution buffered at \(\mathrm{pH}=3.55\) ?

Amino acids are the building blocks for all proteins in our bodies. A structure for the amino acid alanine is All amino acids have at least two functional groups with acidic or basic properties. In alanine, the carboxylic acid group has \(K_{\mathrm{a}}=4.5 \times 10^{-3}\) and the amino group has \(K_{\mathrm{b}}=\) \(7.4 \times 10^{-5}\). Because of the two groups with acidic or basic properties, three different charged ions of alanine are possible when alanine is dissolved in water. Which of these ions would predominate in a solution with \(\left[\mathrm{H}^{+}\right]=1.0 M ?\) In a solution with \(\left[\mathrm{OH}^{-}\right]=1.0 M ?\)

Consider the following four titrations. i. \(100.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{HCl}\) titrated by \(0.10 \mathrm{M} \mathrm{NaOH}\) ii. \(100.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{NaOH}\) titrated by \(0.10 \mathrm{M} \mathrm{HCl}\) iii. \(100.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{2}\) titrated by \(0.10 \mathrm{M} \mathrm{HCl}\) iv. \(100.0 \mathrm{~mL}\) of \(0.10 \mathrm{M}\) HF titrated by \(0.10 \mathrm{M} \mathrm{NaOH}\) Rank the titrations in order of: a. increasing volume of titrant added to reach the equivalence point. b. increasing \(\mathrm{pH}\) initially before any titrant has been added. c. increasing \(\mathrm{pH}\) at the halfway point in equivalence. d. increasing \(\mathrm{pH}\) at the equivalence point. How would the rankings change if \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\) replaced \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) and if \(\mathrm{HOC}_{6} \mathrm{H}_{5}\) replaced \(\mathrm{HF}\) ?
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