(a) Calculate the pH of a buffer that is \(0.125 \mathrm{M}\) in \(\mathrm{NaHCO}_{3}\) and \(0.095 \mathrm{M}\) in $\mathrm{Na}_{2} \mathrm{CO}_{3} .\( (b) Calculate the pH of a solution formed by mixing \)25 \mathrm{~mL}\( of \)0.25 \mathrm{M} \mathrm{NaHCO}_{3}\( with \)75 \mathrm{~mL}$ of \(0.15 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\)

In this part, we have a solution that is \(0.125 \, \mathrm{M}\) in \(\mathrm{NaHCO}_{3}\) and \(0.095\, \mathrm{M}\) in \(\mathrm{Na}_{2}\mathrm{CO}_{3}\). The \(\mathrm{NaHCO}_{3}\) is the weak acid, and \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) is the conjugate base. The acid dissociation constant (\(K_a\)) for \(\mathrm{HCO}_{3}^{-}\) can be found in a reference: \(K_a = 4.3 \times 10^{-7}\) First, we will find the \(pK_a\) using the given \(K_a\): \[pK_a = -\log(K_a) = -\log(4.3 \times 10^{-7})\] Now use the Henderson-Hasselbalch equation to find the pH of the buffer: \[pH = pK_a + \log(\frac{[\mathrm{A}^-]}{[\mathrm{HA}]})\] Plug in the values for \(pK_a\) and the concentrations: \[pH = -\log(4.3 \times 10^{-7}) + \log(\frac{0.095}{0.125})\] Calculate the pH: \[pH \approx 6.12\]

In this part, we have to calculate the pH of the solution formed by mixing \(25 \, \mathrm{mL}\) of \(0.25\, \mathrm{M} \, \mathrm{NaHCO}_{3}\) with \(75\, \mathrm{mL}\) of \(0.15\, \mathrm{M}\, \mathrm{Na}_{2}\mathrm{CO}_{3}\). First, we will find the new concentrations of \(\mathrm{NaHCO}_{3}\) and \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) in the mixed solution: Initial moles of \(\mathrm{NaHCO}_3\): moles = (volume in L) × (molarity) \[\text{moles of }\mathrm{NaHCO}_{3} =0.025\, \text{L} \times 0.25\, \mathrm{M} = 0.00625 \, \text{mol}\] Initial moles of \(\mathrm{Na}_{2}\mathrm{CO}_3\): \[\text{moles of }\mathrm{Na}_{2}\mathrm{CO}_{3} = 0.075\, \text{L} \times 0.15\, \mathrm{M} = 0.01125\, \text{mol}\] Total volume of the mixture: \[V_{total} = 0.025\, \text{L} + 0.075\, \text{L} = 0.1\, \text{L}\] New concentrations: \[[\mathrm{HCO}_{3}^{-}] =\frac{0.00625\, \text{mol}}{0.1\, \text{L}} = 0.0625 \, \mathrm{M}\] \[[\mathrm{CO}_{3}^{2-}] =\frac{0.01125\, \text{mol}}{0.1\, \text{L}} = 0.1125\, \mathrm{M}\] Now, we can use the Henderson-Hasselbalch equation again, plugging in the new concentrations: \[pH = pK_a + \log(\frac{[\mathrm{A}^-]}{[\mathrm{HA}]})\] \[pH = -\log(4.3 \times 10^{-7}) + \log(\frac{0.1125}{0.0625})\] Calculate the pH: \[pH \approx 6.33\]