Calculate the number of moles of \(\mathrm{HCl}(g)\) that must be added to 1.0 \(\mathrm{L}\) of 1.0 $\mathrm{M} \mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}$ to produce a solution buffered at each pH. $$ \text{(a)}\mathrm{pH}=\mathrm{p} K_{\mathrm{a}} \quad \text { b. } \mathrm{pH}=4.20 \quad \text { c. } \mathrm{pH}=5.00 $$

The first thing we need to do is determine the pKa of the weak acid. The formula for acetic acid, a common weak acid, is \(HC_2H_3O_2\). The acid dissociation constant, \(K_a\), for acetic acid is approximately \(1.8 \times 10^{-5}\). Using the formula for pKa, we can calculate it as follows: \(pK_a = -\log{K_a}\) \(pK_a = -\log{(1.8 \times 10^{-5})} \) Using a calculator, we get: \(pK_a \approx 4.74\) Now, we will use the pKa value to determine the required moles of \(HCl(g)\) for each specified pH value.

At pH = pKa, the buffer capacity is at its maximum. In this case: \(pH = pK_a\) \(4.74 = 4.74 + \log{\frac{[C_2H_3O_2^-]}{[HC_2H_3O_2]}}\) Since the left and right sides of the equation are equal, we get: \(0 = \log{\frac{[C_2H_3O_2^-]}{[HC_2H_3O_2]}}\) Raising both sides as exponential, we have: \(1 = \frac{[C_2H_3O_2^-]}{[HC_2H_3O_2]}\) Because 1.0 L of 1.0 M NaC2H3O2 is used, there are initially 1 mole of \(C_2H_3O_2^-\) and 0 moles of \(HC_2H_3O_2\). For the solution to be buffered at pH = pKa, we need an equal number of moles of \(HC_2H_3O_2\) to \(C_2H_3O_2^-\). Since the reaction consumes one mole of \(C_2H_3O_2^-\) and produces one mole of \(HC_2H_3O_2\), adding x moles of \(HCl(g)\) gives: \(x = 1 - x\) \(2x = 1\) \(x = \frac{1}{2}\) Hence, at pH = pKa, we need to add 0.5 moles of \(HCl(g)\).

For pH = 4.20, we have: \(4.20 = 4.74 + \log{\frac{[C_2H_3O_2^-]}{[HC_2H_3O_2]}}\) Rearranging the equation to solve for the ratio of acid/base: \(\frac{[C_2H_3O_2^-]}{[HC_2H_3O_2]} = 10^{-0.54}\) Now, let the number of moles of \(HCl(g)\) added be x. Then: \(\frac{1-x}{x} = 10^{-0.54}\) Rearranging and solving for x: \(x = \frac{1}{10^{-0.54} + 1}\) Using a calculator: \(x \approx 0.361\) Thus, to get a pH of 4.20, we need to add approximately 0.361 moles of \(HCl(g)\).

Finally, for pH = 5.00, we have: \(5.00 = 4.74 + \log{\frac{[C_2H_3O_2^-]}{[HC_2H_3O_2]}}\) Rearranging the equation to solve for the ratio of acid/base: \(\frac{[C_2H_3O_2^-]}{[HC_2H_3O_2]} = 10^{0.26}\) Now, let the number of moles of \(HCl(g)\) added be x. Then: \(\frac{1-x}{x} = 10^{0.26}\) Rearranging and solving for x: \(x = \frac{1}{10^{0.26} + 1}\) Using a calculator: \(x \approx 0.212\) Thus, to get a pH of 5.00, we will need to add approximately 0.212 moles of \(HCl(g)\).