A 1.50 L buffer solution is 0.250 M in HF and 0.250 M in NaF.Calculate the pH of the solution after the addition of 0.0500 moles of solid NaOH.Assume no volume change upon the addition of base.The K_a for HF is 3.5 × 10^-4. A) 3.34 B) 3.46 C) 3.57 D) 3.63 E) 2.89

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The addition of NaOH to the buffer solution will react with HF to form NaF and water. Since 0.0500 moles of NaOH are added, the concentration of HF will decrease by 0.0500 moles/L, and the concentration of NaF will increase by the same amount. The new concentrations in the 1.50 L solution are:- HF: $0.250 - \frac{0.0500}{1.50} = 0.2167$ M- NaF: $0.250 + \frac{0.0500}{1.50} = 0.2833$ MUsing the Henderson-Hasselbalch equation: $pH = pKa + \log\left(\frac{[A^-]}{[HA]}\right)$Given $pKa = -\log(3.5 \times 10^{-4}) = 3.46$,$pH = 3.46 + \log\left(\frac{0.2833}{0.2167}\right) = 3.46 + \log(1.307) = 3.46 + 0.116 = 3.576$Rounded to two decimal places, the pH is approximately 3.57.

A 150 L Buffer Solution Is 0