Question

1. a solution is 0.045 m h₂c₂o₄. determine the ph of the solution and c₂o₄²⁻.

Explanation:

Step1: Write the dissociation of $H_2C_2O_4$

$H_2C_2O_4
ightleftharpoons H^++HC_2O_4^-$ with $K_{a1} = 5.9\times 10^{-2}$; $HC_2O_4^-
ightleftharpoons H^++C_2O_4^{2 -}$ with $K_{a2}=6.4\times 10^{-5}$
Since $K_{a1}\gg K_{a2}$, the first - step dissociation mainly determines the $[H^+]$ concentration. Let $x$ be the concentration of $H^+$ and $HC_2O_4^-$ ions at equilibrium from the first - step dissociation. Then $[H_2C_2O_4]=0.045 - x$, $[H^+]=x$, $[HC_2O_4^-]=x$.
$K_{a1}=\frac{[H^+][HC_2O_4^-]}{[H_2C_2O_4]}=\frac{x\cdot x}{0.045 - x}=5.9\times 10^{-2}$
Assume $0.045 - x\approx0.045$ (since $K_{a1}$ is not extremely small, we can check the assumption later). Then $x^2 = 5.9\times 10^{-2}\times0.045$, $x=\sqrt{5.9\times 10^{-2}\times0.045}\approx0.0163$ M.
Check the assumption: $\frac{0.0163}{0.045}\times100\%\approx36.2\%>5\%$, so we need to solve the quadratic equation $5.9\times 10^{-2}=\frac{x^2}{0.045 - x}$, $x^2+5.9\times 10^{-2}x-(5.9\times 10^{-2}\times0.045)=0$.
Using the quadratic formula $x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$ for the quadratic equation $ax^2+bx + c = 0$. Here $a = 1$, $b = 5.9\times 10^{-2}$, $c=-(5.9\times 10^{-2}\times0.045)$.
$x=\frac{-5.9\times 10^{-2}\pm\sqrt{(5.9\times 10^{-2})^2-4\times1\times(-5.9\times 10^{-2}\times0.045)}}{2\times1}\approx0.015$ M. So $[H^+]\approx0.015$ M.

Step2: Calculate the pH

$pH=-\log[H^+]=-\log(0.015)\approx1.82$

Step3: Calculate $[C_2O_4^{2 -}]$

From the second - step dissociation $HC_2O_4^-
ightleftharpoons H^++C_2O_4^{2 -}$, $K_{a2}=\frac{[H^+][C_2O_4^{2 -}]}{[HC_2O_4^-]}$. Since from the first - step dissociation $[H^+]\approx[HC_2O_4^-]$, then $[C_2O_4^{2 -}]=K_{a2}=6.4\times 10^{-5}$ M

Answer:

$pH\approx1.82$, $[C_2O_4^{2 -}]=6.4\times 10^{-5}$ M