## General Chemistry: Principles and Modern Applications (10th Edition)

$[H_3O^+] = 0.21 \space M$ $[HS{O_4}^-] = 0.19 \space M$ $[S{O_4}^{2-}] =0.011 \space M$
1. Draw the ICE table for the first and second reactions: $$\begin{vmatrix} Compound& [ H_2SO_4 ]& [ HS{O_4}^- ]& [ H_2O ]\\ Initial& 0.20 & 0 & 0 \\ Change& -0.20& +0.20& +0.20\\ Equilibrium& 0& 0.20& 0.20 \\ \end{vmatrix}$$ Since $H_2SO_4$ is a strong acid, it will react completely, so x = 0.20 $$\begin{vmatrix} Compound& [ HS{O_4}^- ]& [ S{O_4}^{2-} ]& [ H_3O^+ ]\\ Initial& 0.20 & 0 & 0.20 \\ Change& -x& +x& +x\\ Equilibrium& 0.20 -x& 0 +x& 0.20 +x\\ \end{vmatrix}$$ 2. Write the expression for $K_a$, and substitute the concentrations: - The exponent of each concentration is equal to its balance coefficient. $$K_a = \frac{[Products]}{[Reactants]} = \frac{[ S{O_4}^{2-} ][ H_3O^+ ]}{[ HS{O_4}^- ]}$$ $$K_a = \frac{(x)( 0.20 + x)}{0.20 - x}$$ 3. Assuming $0.20 \gt\gt x:$ $$K_a = \frac{(x)( 0.20 )}{[ HS{O_4}^- ]_{initial}}$$ $$x = \frac{K_a \times [ HS{O_4}^- ]_{initial}}{ 0.20 } = \frac{ 1.1 \times 10^{-2} \times 0.20 }{ 0.20 }$$ $x = 0.011$ 4. Test if the assumption was correct: $$\frac{ 0.011 }{ 0.20 } \times 100\% = 5.5 \%$$ The percent is approximately 5%, therefore, the difference between this and the actual number is not significant for a 2 significant figures sum. 5. Determine the hydronium concentration: $$[H_3O^+] = x + [H_3O^+]_{initial} = 0.011 + 0.20 = 0.21$$ $[H_3O^+] = 0.21 \space M$ $[HS{O_4}^-] = 0.20 - 0.011 = 0.19 \space M$ $[S{O_4}^{2-}] = x = 0.011 \space M$