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. a. \(\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}\) b. \(\mathrm{pH}=4.20\) c. \(\mathrm{pH}=5.00\)

The \(K_{a}\) and \(\mathrm{p}K_{\mathrm{a}}\) of acetic acid are given as: \[K_{a} = 1.8 \times 10^{-5}\] \[\mathrm{p}K_{\mathrm{a}} = -\log K_{\mathrm{a}} = -\log(1.8 \times 10^{-5}) = 4.74\]

The Henderson-Hasselbalch equation is given by: \[{\mathrm{pH}}={\mathrm{p}K_{\mathrm{a}}} + \log\left(\frac{\left[\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_{2}^{-}\right]}{\left[\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\right]}\right)\] Rearrange the equation to find the ratio: \[\left[\frac{\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_{2}^{-}}{\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}}\right] = 10^{\mathrm{(pH - pK}_{a}\mathrm{)}\mathrm{)}}\] Now, we will use this equation to find the ratio for each given pH value. a. \(\mathrm{pH}=\mathrm{p} K_{\mathrm{a}} = 4.74\) \[\left[\frac{\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_{2}^{-}}{\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}}\right]_{a} = 10^{(4.74-4.74)} = 10^0 = 1\] b. \(\mathrm{pH}=4.20\) \[\left[\frac{\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_{2}^{-}}{\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}}\right]_{b} = 10^{(4.20-4.74)} = 10^{-0.54} \approx 0.287\] c. \(\mathrm{pH}=5.00\) \[\left[\frac{\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_{2}^{-}}{\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}}\right]_{c} = 10^{(5.00-4.74)} = 10^{0.26} \approx 1.819\]

Since we have a \(1.0 \mathrm{~L}\) solution of \(1.0 \mathrm{M} \mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\), initially, we have 1 mole of \(\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_{2}^{-}\) and 0 moles of \(\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\). a) For pH \(= 4.74\), we need an equal amount of \(\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\), which is 1 mole. Adding 1 mole of \(\mathrm{HCl}\) to 1 mole of acetate ions gives us 1 mole of acetic acid, as they react with a 1:1 ratio. b) For pH \(= 4.20\), we need a \(\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\) to \(\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_{2}^{-}\) ratio of 0.287. Let's use x to represent the moles of \(\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\): \[\frac{x}{1.0-x} = 0.287\] \[x = 0.287(1.0-x)\] \[x = 0.287 - 0.287x\] \[1.287x = 0.287\] \[x \approx 0.223\] So, we need to add approximately 0.223 moles of \(\mathrm{HCl}\). c) For pH \(= 5.00\), we need a \(\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\) to \(\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_{2}^{-}\) ratio of 1.819. \[\frac{x}{1.0-x} = 1.819\] \[x = 1.819(1.0-x)\] \[x = 1.819 - 1.819x\] \[2.819x = 1.819\] \[x \approx 0.645\] So, we need to add approximately 0.645 moles of \(\mathrm{HCl}\). In conclusion, we need to add: - For pH \(= 4.74\), 1 mole of \(\mathrm{HCl}\). - For pH \(= 4.20\), approximately 0.223 moles of \(\mathrm{HCl}\). - For pH \(= 5.00\), approximately 0.645 moles of \(\mathrm{HCl}\).