Question

Citric acid (H₃C₆H₅O₇) is a triprotic acid with Ka₁ = 8.4 × 10^-4, Ka₂= 1.8 × 10^-5, and Ka₃ = 4.0 × 10^-6. Calculate the pH of 0.15 M citric acid.

Short answer

pH of the given solution is 1.95

Step-by-step solution

Polyprotic weak acids

For weak polyprotic acids successive K_a values are so much smaller than the first value that only the first dissociation step makes a significant contribution to the equilibrium concentration of H⁺ ions. So, for the calculation of pH only first dissociation step is considered.

Calculation of pH

Citric acid dissociates as:

$\begin{array}{l}\begin{array}{ccccc}& {\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}& \rightleftarrows & {\text{H}}^{\text{+}}\text{+}& {\text{H}}_{\text{2}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{{\text{O}}_{\text{7}}}^{\text{-}}\\ \text{Initial}& \text{0.15}& & \text{0}& \text{0}\\ \text{change}& \text{-x}& & \text{+x}& \text{+x}\\ \text{At\hspace{0.17em}Equil.}& \text{0.15-x}& & \text{x}& \text{x}\end{array}\\ \end{array}$

$\begin{array}{l}{\text{K}}_{\text{a1}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0.15-x}}{\text{=8.4×10}}^{\text{-4}}\\ {{\text{x=1.12×10}}^{\text{-2}}}^{}\end{array}$

$\begin{array}{l}{\text{pH=-log[H}}^{\text{+}}\text{]}\\ {\text{pH=-log(1.12×10}}^{\text{-2}}\text{)}\\ \text{pH=1.95}\end{array}$

Hence, pH of the given solution comes out to be 1.95.