A buffer is made using \(45.0 \mathrm{mL}\) of \(0.750 \mathrm{M} \mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\left(K_{\mathrm{a}}=\right.\) \(1.3 \times 10^{-5}\) ) and \(55.0 \mathrm{mL}\) of \(0.700 \mathrm{M} \mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{2} .\) What volume of 0.10 \(M\) NaOH must be added to change the pH of the original buffer solution by \(2.5 \% ?\)

To find the initial pH, we will use the Henderson-Hasselbalch equation, which is given by: pH = pKa + log(\(\frac{[A^-]}{[HA]}\)) where pKa = -log(Ka) and [A^-] is the concentration of the conjugate base (NaC3H5O2) and [HA] is the concentration of the weak acid (HC3H5O2). First, let's find the pKa from the given Ka value: pKa = -log(\(1.3 \times 10^{-5}\)) = 4.89 Now we will find the initial concentrations of the weak acid and the conjugate base: [HA] = (0.750 M * 45.0 mL) / (45.0 mL + 55.0 mL) = 0.403 M [A^-] = (0.700 M * 55.0 mL) / (45.0 mL + 55.0 mL) = 0.350 M Substitute the values into the Henderson-Hasselbalch equation: Initial pH = 4.89 + log(\(\frac{0.350}{0.403}\)) ≈ 4.79

We are given that the pH of the buffer solution needs to be changed by 2.5%. To determine the change in pH required, we can calculate it as follows: Change in pH = Initial pH * 0.025 = 4.79 * 0.025 ≈ 0.12

The neutralization reaction that occurs when NaOH is added to the buffer solution is: HC3H5O2 + OH¯ → H2O + C3H5O2¯ Since the change in pH is small, we can assume that for every mole of OH¯ added, one mole of HC3H5O2 will be converted to C3H5O2¯. So, the moles of OH¯ needed for the desired change in pH can be calculated using the initial and final concentrations of HC3H5O2 and C3H5O2¯: Initial moles of HC3H5O2 = 0.403 M * (45.0 mL + 55.0 mL) = 0.403 mol/L * 0.1 L = 0.0403 mol Final moles of HC3H5O2 = 0.0403 mol - n (where n is the moles of OH¯ added) Final moles of C3H5O2¯ = 0.350 M * (45.0 mL + 55.0 mL) + n = 0.350 mol/L * 0.1 L + n = 0.035 + n Substitute the final concentrations into the Henderson-Hasselbalch equation and solve for n: Initial pH + change in pH = pKa + log(\(\frac{0.035 + n}{0.0403 - n}\)) 4.79 + 0.12 = 4.89 + log(\(\frac{0.035 + n}{0.0403 - n}\)) 0.02 = log(\(\frac{0.035 + n}{0.0403 - n}\)) \(10^{0.02}\) = \(\frac{0.035 + n}{0.0403 - n}\) 0.0403\(10^{0.02}\) - n\(10^{0.02}\) = 0.035 + n Now, we can solve for n: n ≈ 0.000599 mol (3 significant figures)

Now we can calculate the volume of 0.10 M NaOH required to achieve the desired change in pH: Volume of NaOH = (moles of OH¯) / (concentration of NaOH) = 0.000599 mol / 0.10 M ≈ 5.99 mL So, approximately 5.99 mL of 0.10 M NaOH must be added to change the pH of the original buffer solution by 2.5%.