A buffer is made up of \(0.300\) L each of \(0.500 \mathrm{MKH}_{2} \mathrm{PO}_{4}\) and \(0.317 \mathrm{M}\) \(\mathrm{K}_{2} \mathrm{HPO}_{4}\). Assuming that volumes are additive, calculate (a) the \(\mathrm{pH}\) of the buffer. (b) the \(\mathrm{pH}\) of the buffer after the addition of \(0.0500 \mathrm{~mol}\) of \(\mathrm{HCl}\) to \(0.600 \mathrm{~L}\) of buffer. (c) the \(\mathrm{pH}\) of the buffer after the addition of \(0.0500 \mathrm{~mol}\) of \(\mathrm{NaOH}\) to \(0.600 \mathrm{~L}\) of buffer.

To calculate the initial pH of the buffer, we need to apply the Henderson-Hasselbalch equation: \(\mathrm{pH} = \mathrm{p}K_{a} + \log \frac{\left[\mathrm{A}^{-}\right]}{\left[\mathrm{HA}\right]}\). In this case, our acid is \(\mathrm{KH}_{2}\mathrm{PO}_{4}\) and our base is \(\mathrm{K}_{2}\mathrm{HPO}_{4}\). First, we need to find the \(\mathrm{p}K_{a}\) value for \(\mathrm{KH}_{2}\mathrm{PO}_{4}\). This value is given to be \(7.2\). Now, we need to find the concentrations of the acid and the base. The initial concentrations are as follows: \(\left[\mathrm{KH}_{2}\mathrm{PO}_{4}\right] = 0.5 \mathrm{M}\) and \(\left[\mathrm{K}_{2}\mathrm{HPO}_{4}\right] = 0.317 \mathrm{M}\) Using the Henderson-Hasselbalch equation: \(pH = 7.2 + \log\left(\frac{0.317}{0.5}\right)\) Calculating this gives us the initial pH of the buffer solution, which is approximately \(\sim 6.86\).

Adding \(\mathrm{HCl}\) to the buffer will cause the acid component to increase in concentration. We are given that \(0.0500 \; \mathrm{mol}\) of \(\mathrm{HCl}\) is added to \(0.600 \; \mathrm{L}\) of buffer. To find the new concentrations of the acid and base, we will use the following relations: - The amount of acid will increase by the moles of HCl added - The amount of base will decrease by the same amount New concentration of \(\mathrm{KH}_{2}\mathrm{PO}_{4}\) = \(\frac{0.5 \cdot 0.3 + 0.0500}{0.600} = 0.558 \; \mathrm{M}\) New concentration of \(\mathrm{K}_{2}\mathrm{HPO}_{4}\) = \(\frac{0.317 \cdot 0.3 - 0.0500}{0.600} = 0.259 \; \mathrm{M}\) Now, we can use the Henderson-Hasselbalch equation again: \(pH = 7.2 + \log\left(\frac{0.259}{0.558}\right)\) Calculating this gives us the pH of the buffer solution after the addition of \(\mathrm{HCl}\), which is approximately \(\sim 6.75\).

Adding \(\mathrm{NaOH}\) to the buffer will cause the base component to increase in concentration. We are given that \(0.0500 \; \mathrm{mol}\) of \(\mathrm{NaOH}\) is added to \(0.600 \; \mathrm{L}\) of buffer. To find the new concentrations of the acid and base, we will use the following relations: - The amount of acid will decrease by the moles of NaOH added - The amount of base will increase by the same amount New concentration of \(\mathrm{KH}_{2}\mathrm{PO}_{4}\) = \(\frac{0.5 \cdot 0.3 - 0.0500}{0.600} = 0.442 \; \mathrm{M}\) New concentration of \(\mathrm{K}_{2}\mathrm{HPO}_{4}\) = \(\frac{0.317 \cdot 0.3 + 0.0500}{0.600} = 0.376 \; \mathrm{M}\) Now, we can use the Henderson-Hasselbalch equation again: \(pH = 7.2 + \log\left(\frac{0.376}{0.442}\right)\) Calculating this gives us the pH of the buffer solution after the addition of \(\mathrm{NaOH}\), which is approximately \(\sim 7.00\). In summary, the initial pH of the buffer is \(\sim 6.86\), the pH after the addition of \(\mathrm{HCl}\) is \(\sim 6.75\), and the pH after the addition of \(\mathrm{NaOH}\) is \(\sim 7.00\).