Question 18.99: Compare the contribution of each acid by calculating the con......

M\text{ H}_2\text{O}_2=\left\lgroup\frac{1.00\text{ g}}{\text{mL}} \right\rgroup \left\lgroup\frac{3\%\text{ H}_2\text{O}_2}{100\%} \right\rgroup \left\lgroup\frac{1\text{ mol H}_2\text{O}_2}{34.02\text{ g H}_2\text{O}_2} \right\rgroup \left\lgroup\frac{1\text{ mL}}{10^{-3}\text{ L}} \right\rgroup=0.881834 \ M\text{ H}_2\text{O}_2\text{ (unrounded)} \\ \text{Find }K_\text{a}\text{ from p}K_\text{a}: K_\text{a} = 10^{-\text{p}K\text{a}} = 10^{-11.75} = 1.778279 \times 10^{-12} (unrounded)

K_\text{a}=1.778279\times 10^{-12}=\frac{[\text{H}_3\text{O}^+][\text{HO}_2^-]}{[\text{H}_2\text{O}_2]} \\ K_\text{a}=1.778279\times 10^{-12}=\frac{(\text{x})(\text{x})}{(0.881834-\text{x})} Assume x is small compared to 0.881834.

Check assumption: ( 1.2522567 \times 10^{-6} /0.881834) \times 100\% = 0.0001\% error, so the assumption is valid.

From Appendix C, K_\text{a}\text{ for phosphoric acid is }7.2 \times 10^{-3}. The subsequent K_\text{a} values may be ignored. In this calculation x is not negligible since the initial concentration of acid is less than the K_\text{a}.

The problem will need to be solved as a quadratic.
\text{x}^2 = (7.2 \times 10^{-3}) ( 1.0205 \times 10^{-4} – \text{x}) = 7.3476 \times 10^{-7}- 7.2 \times 10^{-3} \text{ x} \\ \text{x}^2 + 7.2 \text{ x}\times 10^{-3}\text{ x} – 7.3476 \times 10^{-7} = 0 \\  \text{a} = 1\quad \text{b} = 7.2 \times 10^{-3} \quad \text{c} = -7.3476 \times 10^{-7}\\ \text{x}=\frac{-7.2\times 10^{-3}\pm \sqrt{(7.2\times 10^{-3})^2-4(1)(-7.3476\times 10^{-7})} }{2(1)} \\ \text{x}=1.00643\times 10^{-4} \ M\text{ H}_3\text{O}^+\text{ (unrounded)}

The concentration of hydronium ion produced by the phosphoric acid, 1 \times 10^{-4} \ M, is greater than the concentration produced by the hydrogen peroxide, 1 \times 10^{-6} \ M. Therefore, the phosphoric acid contributes more \text{H}_3\text{O}^+ to the solution.