[tex]\text{ HCOOH + H}_2\text{O }\leftrightarrow\text{ HCOO}^-\text{ + H}_3\text{O}^+\text{ Ka=1.8}\times10^{-4}[/tex][tex]\text{ Ka = }\frac{\lbrack\text{ HCOO}^-\rbrack\lbrack\text{ H}_3\text{O}^+\rbrack}{\lbrack\text{ HCOOH}\rbrack}=\text{ }\frac{\text{ x}^2}{0.025\text{ - x}}\text{ }[/tex]
We can neglect change in X, so
[tex]\text{ Ka = }\frac{\text{ x}^2}{0.025}[/tex][tex]\text{ x = }\sqrt{\text{ Ka }\times0.025}=\text{ }\sqrt{1.8\times10^{-4}\times0.025}=\text{ 0.002 M}[/tex]
Now the equilibrum of hypobromous acid (smaller Ka)
[tex]\text{ HBrO + H}_2\text{O }\leftrightarrow\text{ BrO}^-\text{ + H}_3\text{O}^+\text{ Ka= 2.0}\times10^{-9}[/tex]
[tex]\text{ Ka = }\frac{\lbrack\text{ BrO}^-\rbrack\lbrack\text{ H}_3\text{O}^+\rbrack}{\lbrack\text{ HBrO}\rbrack}=\text{ }\frac{\text{ x \lparen x+0.002\rparen}}{0.05\text{ -x}}\text{ }[/tex]
We can neglect change in X, so
[tex]\text{ Ka = }\frac{\text{ x}^2\text{ + 0.002x}}{0.05}[/tex]
Solve for x
X is very small, order of magnitude -8. Then it can be neglected
So [H3O+] = 0.002 M pH= -log(0.002) = 2.70
