Question

How to calculate the pH of a buffer solution1) Calculate the pH of a solution prepared by dissolving 2.35 g of sodium acetate, CH3COONa, in 50.0 mL of 0.20 Maceticacid, CH3COOH(aq). Assume the volume change upon dissolving the sodium acetate is negligible. K₂ of CH₂COOH is1.75 X 10.5.pH =

Answer 1

The equlibrium that exist is:

$\begin{gathered} CH_3COOH\rightleftarrows CH_3COO^-+H^+ \\ CH_3COONa\rightleftarrows CH_3COO^-+Na^+ \end{gathered}$

-We will determine the concentration of sodium acetate:

$\begin{gathered} moles\text{ }of\text{ }CH_3COONa=\frac{mass}{molar\text{ }mass} \\ \\ moles\text{ }of\text{ }CH_3COONa=\frac{2.35g}{82.03gmol^{-1}} \\ \\ moles\text{ }of\text{ }CH_3COONa=0.02865mole \\ \\ Concentration\text{ }of\text{ }CH_3COONa=\frac{moles}{volume(L)} \\ \\ Concentration\text{ }of\text{ }CH_3COONa=\frac{0.02865\text{ }mole}{0.05L} \\ \\ Concentration\text{ }of\text{ }CH_3COONa=0.573mol•L^{-1} \end{gathered}$

Now that we have the concentration of the conjugate base we can use Henderson-Hasselbach equation to determine the pH:

$\begin{gathered} pH=-logK_a+log\frac{[conjugate\text{ }base]}{[acid]} \\ \\ pH=-log1.75\times10^{-5}+log\frac{[0.573]}{[0.2]} \\ pH=4.76+0.46 \\ pH=5.22 \end{gathered}$

Answer: The pH of the solution is 5.22,