Question

suppose a 0.010 m aqueous solution of sulfuric acid (h₂so₄) is prepared. calculate the equilibrium molarity of so₄²⁻. youll find information on the properties of sulfuric acid in the aleks data resource. round your answer to 2 significant digits.

Explanation:

Step1: Consider sulfuric acid dissociation

Sulfuric acid ($H_2SO_4$) is a strong - acid. The first dissociation is complete: $H_2SO_4
ightarrow H^++HSO_4^-$. For a $0.010M$ $H_2SO_4$ solution, after the first dissociation, $[HSO_4^-]=0.010M$ and $[H^+]=0.010M$. The second dissociation of $HSO_4^-$ is partial: $HSO_4^-
ightleftharpoons H^++SO_4^{2 - }$, and the acid - dissociation constant $K_{a2}$ of $HSO_4^-$ is $1.2\times10^{-2}$.

Step2: Set up the expression for $K_{a2}$

Let $x = [SO_4^{2 - }]$, then $[H^+]=0.010 + x$ and $[HSO_4^-]=0.010 - x$. The expression for $K_{a2}$ is $K_{a2}=\frac{[H^+][SO_4^{2 - }]}{[HSO_4^-]}$, so $1.2\times10^{-2}=\frac{(0.010 + x)x}{0.010 - x}$.

Step3: Make an approximation (if valid)

Since $K_{a2}=1.2\times10^{-2}$, the approximation $0.010 + x\approx0.010$ and $0.010 - x\approx0.010$ may not be valid. We solve the quadratic equation $1.2\times10^{-2}(0.010 - x)=(0.010 + x)x$. Expanding gives $1.2\times10^{-4}-1.2\times10^{-2}x = 0.010x+x^{2}$. Rearranging to the standard quadratic - form $x^{2}+0.022x - 1.2\times10^{-4}=0$.

Step4: Solve the quadratic equation

Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$. Here, $a = 1$, $b = 0.022$, and $c=-1.2\times10^{-4}$. First, calculate the discriminant $\Delta=b^{2}-4ac=(0.022)^{2}-4\times1\times(-1.2\times10^{-4})=4.84\times10^{-4}+4.8\times10^{-4}=9.64\times10^{-4}$. Then $x=\frac{-0.022\pm\sqrt{9.64\times10^{-4}}}{2}=\frac{-0.022\pm0.031}{2}$. We take the positive root $x=\frac{-0.022 + 0.031}{2}=4.5\times10^{-3}M$.

Answer:

$4.5\times10^{-3}M$