Question

Calculate the pH of a solution that is 0.050 M in hypobromous acid and 0.025 M in formic acid.

Answer 1

`\text{ HCOOH + H}_2\text{O }\leftrightarrow\text{ HCOO}^-\text{ + H}_3\text{O}^+\text{ Ka=1.8}*10^(-4)`
`\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{ }`

We can neglect change in X, so

`\text{ Ka = }\frac{\text{ x}^2}{0.025}`
`\text{ x = }\sqrt{\text{ Ka }*0.025}=\text{ }\sqrt{1.8*10^(-4)*0.025}=\text{ 0.002 M}`

Now the equilibrum of hypobromous acid (smaller Ka)

`\text{ HBrO + H}_2\text{O }\leftrightarrow\text{ BrO}^-\text{ + H}_3\text{O}^+\text{ Ka= 2.0}*10^(-9)`

`\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{ }`

We can neglect change in X, so

`\text{ Ka = }\frac{\text{ x}^2\text{ + 0.002x}}{0.05}`

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