Question

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

Short answer

Answer: (a) The initial pH of the buffer is 7.01. (b) The pH after the addition of HCl is 5.94. (c) The pH after the addition of NaOH is 7.01.

Step-by-step solution

Calculate initial concentrations of weak acid and weak base

Since the volume of each solution is the same (\(0.300\) L), the total volume of the buffer is \(0.300\,\text{L} + 0.300\,\text{L} = 0.600\,\text{L}\). Next, we calculate the new concentrations of the weak acid and weak base in the buffer by using the dilution equation \(c_1V_1=c_2V_2\). So, \(c_2 = \frac{c_1V_1}{V_2}\). For \(KH_2PO_4\): \( [\mathrm{KH}_{2} \mathrm{PO}_{4}] = \frac{0.500\,\text{M} × 0.300\,\text{L}}{0.600\,\text{L}} = 0.250\,\text{M}\) For \(K_2HPO_4\): \( [\mathrm{K}_{2} \mathrm{HPO}_{4}] = \frac{0.317\,\text{M} × 0.300\,\text{L}}{0.600\,\text{L}} = 0.158\,\text{M}\)

Calculate the initial pH of the buffer

Using the Henderson-Hasselbalch equation: \(\mathrm{pH} = \mathrm{p}K_a + \log \frac{[\mathrm{conj.~base}]}{[\mathrm{acid}]}\) \(K_a\) for \(H_2PO_4^-\) is given as \(6.2 × 10^{-8}\). Thus, \(\mathrm{p}K_a = -\log(6.2 × 10^{-8}) = 7.21\) Now, substitute the values into the equation: \(\mathrm{pH} = 7.21 + \log \frac{0.158}{0.250} = 7.21 - 0.200 = 7.01\) The initial pH of the buffer is 7.01.

Calculate pH after addition of HCl

When HCl is added to the buffer, it reacts with the weak base (\(K_2HPO_4\)). Calculate the new concentrations after adding \(0.0500\,\text{mol}\) of HCl: Moles of weak base: \((0.158\,\text{M})(0.600\,\text{L})=0.0948\,\text{mol}\) Moles of weak acid: \((0.250\,\text{M})(0.600\,\text{L})=0.150\,\text{mol}\) \(0.0500\,\text{mol}\) HCl consumes \(0.0500\,\text{mol}\) of \(K_2HPO_4\) and forms an equal amount of \(KH_2PO_4\). So, the new concentrations are: \([K_2HPO_4] = \frac{0.0948\,\text{mol} - 0.0500\,\text{mol}}{0.600\,\text{L}} = 0.0750\,\text{M}\) \([KH_2PO_4] = \frac{0.150\,\text{mol} + 0.0500\,\text{mol}}{0.600\,\text{L}} = 0.333\,\text{M}\) Now, calculate the new pH: \(\mathrm{pH} = 7.21 + \log \frac{0.0750}{0.333} = 7.21 - 1.27 = 5.94\) The pH after addition of HCl is 5.94.

Calculate pH after addition of NaOH

When NaOH is added to the buffer, it reacts with the weak acid (\(KH_2PO_4\)). Calculate the new concentrations after adding \(0.0500\,\text{mol}\) of NaOH: Moles of weak base: \((0.0750\,\text{M})(0.600\,\text{L})=0.0450\,\text{mol}\) Moles of weak acid: \((0.333\,\text{M})(0.600\,\text{L})=0.200\,\text{mol}\) \(0.0500\,\text{mol}\) NaOH consumes \(0.0500\,\text{mol}\) of \(KH_2PO_4\) and forms an equal amount of \(K_2HPO_4\). So, the new concentrations are: \([K_2HPO_4] = \frac{0.0450\,\text{mol} + 0.0500\,\text{mol}}{0.600\,\text{L}} = 0.158\,\text{M}\) \([KH_2PO_4] = \frac{0.200\,\text{mol} - 0.0500\,\text{mol}}{0.600\,\text{L}} = 0.250\,\text{M}\) Now, calculate the new pH: \(\mathrm{pH} = 7.21 + \log \frac{0.158}{0.250} = 7.21 + (-0.200) = 7.01\) The pH after the addition of NaOH is 7.01. To summarize: (a) The initial pH of the buffer is 7.01. (b) The pH after the addition of HCl is 5.94. (c) The pH after the addition of NaOH is 7.01.