Question

Questions 3-6 refer to rhe solutions below:

Answer 1

Under room temperature where $\text{pK}_w = 14$:

3.) (A), (B), and (E).

4.) (D).

5.) (B).

Explanation

What makes a buffer solution? For a solution to be a buffer, it needs to contain large amounts of a weak acid and its conjugate base ion. Alternatively, the solution may contain large amounts of a weak base and its conjugate acid ion.  

Not every one of the five solutions is a buffer solution.

(A)

Ethanoic acid CH₃COOH (a.k.a. acetic acid) is a weak acid. pKa = 4.756. CH₃COONa is a salt. It dissolves to produce CH₃COO⁻, which is the conjugate base ion of CH₃COOH. The solution in (A) contains equal number of CH₃COOH and CH₃COO⁻, both at 1.0 M.

Refer to the Henderson-Hasselbalch equation for buffers of weak acids.

$\displaystyle \text{pH} = \text{pK}_a + \log{\frac{[\text{Conjugate Ion}]}{[\text{Weak Acid}]}}$.

$\displaystyle \log{\frac{[\text{Conjugate Ion}]}{[\text{Weak Acid}]}} =\ln{1} = 0$.

The pH of the solution in (A) will be the same as the pKa of CH₃COOH. pH = 4.746.

(B)

Consider the hydrogen halides:

• HF: weak acid.
• HCl: strong acid.
• HBr: strong acid.

The radius of halogen atoms increases down the group, and hydrogen-halogen bond becomes weaker. It becomes easier for water to break those bonds. As a result, the strength of hydrogen halides increases down the group. HF is the only weak acid among the common hydrogen halides.

Mixing HBr and KBr at equal ratio will be similar to mixing HCl and KCl at the same ratio. All HBr in the solution breaks down into H⁺ and Br⁻. The pH of the solution will depend only on the concentration of HBr.

$\displaystyle [\text{H}^{+}] = [\text{HBr}] \\\phantom{[\text{H}^{+}]}= \frac{n}{V} \\\phantom{[\text{H}^{+}]}= \frac{c(\text{HBr})\cdot V(\text{HBr})}{V(\text{HBr})+V(\text{KBr})}\\\phantom{[\text{H}^{+}]}=\frac{0.100\;\text{L}\times 1.0\;\text{mol}\cdot\text{L}^{-1}}{0.100\;\text{L}+0.100\;\text{L}} \\\phantom{[\text{H}^{+}]}= 0.50\;\text{mol}\cdot\text{L}^{-1}$.

$\text{pH} = -\log{[\text{H}^{+}] = -\log{0.50} \approx {\bf 0.30}$.

(C)

Similarly to HCl and HBr, HI is also a strong acid. Mixing HI and NaOH at equal ratio will produce a solution of NaI, which is similar to NaCl. The final solution will be neutral. pH = 7 if pKw = 14.

(D)

NH₃ is a weak base. NH₄Cl dissolves completely to produce NH₄⁺ and Cl⁻. NH₄⁺ is the conjugate acid of NH₃. The final solution will contain an equal number of NH₃ and NH₄⁺. pKb = 4.75 for ammonia NH₃.

Apply the Henderson-Hasselbalch equation for buffers of weak bases:

$\displaystyle \textbf{pOH} = \text{pK}_b + \log{\frac{[\text{Conjugate Ion}]}{[\text{Weak Base}]}}= 4.75 + \log{1} = 4.75$.

Note that what this equation gives for buffers of weak bases is the pOH of the solution. pH = pKw - pOH. Assume that pKw = 14. pH = 14 - 4.75 = 9.25.

(E)

The solution in (E) will contain about 1.0 M of CH₃COOH. The volume of the solution will be 200 mL.

$n(\text{CH}_3\text{COO}^{-}) = n(\text{NaOH}] = c\cdot V = 0.10\;\text{mol}$.

$\displaystyle [\text{CH}_3\text{COO}^{-}] = \frac{n}{V} = {0.10}{0.10 + 0.10} = 0.50 \;\text{mol}\cdot\text{L}^{-1}$.

There's nearly no conjugate base of CH₃COOH. As a result, the solution will not be a buffer, and the Henderson-Hasselbalch Equation will not apply. Refer to an ICE table:

$\begin{array}{c|ccccccc}\text{R}&\text{CH}_3\text{COO}^{-} &+&\text{H}_2\text{O}&\rightleftharpoons &\text{CH}_3\text{COOH}&+&\text{OH}^{-}\\\text{I}&0.50\\\text{C}& -x &&&& +x &&+x\\\text{E} &0.50 - x &&&&x&&x\end{array}$

The value of pKa is large. Ka will be small. the value of $x$ will be much smaller than $0.50$ such that $0.50-x \approx 0.50$.

The pKa of a weak acid is the same as pKw divided by the pKb of its conjugate base.

$\displaystyle \frac{[\text{CH}_3\text{COOH}]\cdot[\text{OH}^{-}]}{[\text{CH}_3\text{COO}^{-}]} = \text{K}_b(\text{CH}_3\text{COO}^{-}) \\\phantom{\displaystyle \frac{[\text{CH}_3\text{COOH}]\cdot[\text{OH}^{-}]}{[\text{CH}_3\text{COO}^{-}]} }= \frac{\text{K}_w}{\text{K}_a(\text{CH}_3\text{COOH})} \\\phantom{\displaystyle \frac{[\text{CH}_3\text{COOH}]\cdot[\text{OH}^{-}]}{[\text{CH}_3\text{COO}^{-}]}} = \frac{10^{-14}}{1.75\times 10^{-5}} = 5.71\times 10^{-10}$.

$\displaystyle \frac{x^{2}}{0.50} =5.71\times 10^{-10}$.

$[\text{OH}^{-}] = x \approx 1.69\times 10^{-5}\;\text{mol}\cdot\text{L}^{-1}$.

$\text{pH} = \text{pK}_w + \log{[\text{OH}^{-}]} = 9.23$.