# Question #f096c

Jul 3, 2017

pH = 1.23 with $\left[{H}^{+}\right] \approx 0.0576 M$

#### Explanation:

In polyprotic ionizations such as Phosphoric Acid, it is typically assumed that all of the H^+ ions come from the 1st ionization step. Subsequent ionizations begin with the species' concentrations from the previous ionization step and find that the anionic concentrations equal the acid ionization constants.

For ${H}_{3} P {O}_{4}$:
From 1st Ionization step => ${\left[{H}^{+}\right]}_{1}$ = $\left[{H}_{2} P {O}_{4}^{2 -}\right]$ = $0.0576 M$
From 2nd Ionization step => ${\left[{H}^{+}\right]}_{2} = \left[H P {O}_{4}^{2 -}\right] = 6.2 \times {10}^{-} 8 M$
From 3rd Ionization step => ${\left[{H}^{+}\right]}_{3} = \left[P {O}_{4}^{3 -}\right] = 4.2 \times {10}^{-} 13 M$

$\Sigma \left[{H}^{+}\right] = 0.057600062 M$ which is ~ concentration of ${H}^{+}$ all ionization steps. For calculation of pH, ${H}^{+}$ from ionizations 2 and 3 are negligible.

=> $p H = - \log \left[{H}^{+}\right] = - \log \left(0.0576\right) = 1.23$