Answer
a. Hydronium ion ($H_3O^+$ or $H^+$), $ClO_4^-$ and $H_2O$.
pH = 0.602
b. Hydronium ion ($H_3O^+$ or $H^+$), $NO_3^-$ and $H_2O$.
pH = 0.602
Work Step by Step
1. Identify the reactions that will occur:
- Since $HClO_4$ and $HNO_3$ are strong acids, this reaction should occur completely, and the reactants will be completely consumed:
$HClO_4(aq) + H_2O(l) -- \gt H_3O^+(aq) + ClO_4^-(aq)$
$HNO_3(aq) + H_2O(l) -- \gt H_3O^+(aq) + NO_3^-(aq)$
Therefore, the only compounds that are present in solution at the equilibrium is $H_3O^+$, the conjugate base in each case, and water (because it is the solvent.)
- Since the ratio ($Acid/H_3O^+$) for this reaction is 1 to 1:
$[H_3O^+] = 0.250M$ in both cases.
2. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 0.250)$
$pH = 0.602$
