Question

Phosphoric acid is a common ingredient in traditional cola drinks. It is added to provide the drinks with a pleasantly tart taste. Assuming that in cola drinks the concentration of phosphoric acid is 0.007 M, calculate the pH in this solution.

Short answer

pH is comes out to be 9.8.

Step-by-step solution

H3PO4 Dissociation and [OH-] ion concentration Calculation

The First ionization of H₃PO₄ is represented as:

$\begin{array}{l}{\text{H}}_3{{\text{PO}}_{\text{4}}}_{\left(aq\right)}\rightleftarrows {{\text{\hspace{0.17em}\hspace{1em}H}}^{\text{+}}}_{\left(aq\right)}{\text{+\hspace{0.17em}H}}_{\text{2}}{\text{PO}}_{\text{4}}{{}^{\text{-}}}_{\left(aq\right)}\\ 0.00700\\ 0.007-\text{x\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{1em}\hspace{1em}x\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}x}\end{array}$

$\begin{array}{l}{\text{K}}_{\text{a1}}=\frac{{\text{[H}}^{\text{+}}{\text{][H}}_{\text{2}}{{\text{PO}}_{\text{4}}}^{\text{-}}\text{]}}{{\text{[H}}_3{\text{PO}}_{\text{4}}\text{]}}=\frac{{\text{x}}^2}{0.007-x}=7.5\times {10}^{-3}\\ {\text{x}}^2=7.5\times {10}^{-3}\times 0.007-7.5\times {10}^{-3}\text{x}\\ {\text{x}}^2+7.5\times {10}^{-3}\text{x}-5.25\times {10}^{-5}=0\end{array}$

$\begin{array}{l}\text{For\hspace{0.17em}solving}\\ \frac{-\text{b}\pm \sqrt{{\text{b}}^2-4\text{ac}}}{2a}=\\ \frac{-7.5\times {10}^{-3}\pm \sqrt{{(7.5\times {10}^{-3})}^2-4\left(1\right)(-5.25\times {10}^{-5}})}{2\left(1\right)}\\ \text{x}=6.5\times {10}^{-5}\end{array}$

pH Calculation

$\begin{array}{l}{\text{pOH=\hspace{0.17em}-log[OH}}^{\text{-}}\text{]}\\ {\text{pOH=-log[6.5×10}}^{\text{-5}}\text{]}\\ \text{pOH=4.2}\\ \text{pH=14-pOH=14-4.2=9.8}\end{array}$