Question

What is the pH of 1.30 M Cu(NO3)2 (Ka = 3.00x10^-8)?

Answer 1

`\text{ Cu\lparen NO}_3)_2\text{ }\rightarrow\text{ Cu}^(2+)\text{ + 2NO}_3^-`

The NO3- ion is the conjugate base of a strong acid, it will not affecy the pH, but Cu2+ ion is not conjugate acid of a strong base, it will affect the pH.

`\text{ Cu}^(2+)\text{ + 2H}_2\text{O }\rightleftarrows\text{ Cu\lparen OH\rparen}_2\text{ + 2H}^{+\text{ }}\text{ Ka=3.00x10}^(-8)^{\frac{}{}}`
`\text{ Ka= }\frac{\lbrack\text{ H}^+\rbrack^2}{\lbrack\text{ Cu}^(2+)\rbrack}`

we can use the approximation that [Cu2+] = initial concentration of Cu(NO3)2 because the is Ka is very small

Solve for [H+]:

`\begin{gathered} \lbrack\text{ H}^+\rbrack\text{ = }\sqrt{\text{ Ka }*\text{ }\lbrack\text{ Cu}^(2+)\rbrack} \\ \lbrack\text{ H}^+\rbrack\text{ = }\sqrt{3.00*10^(-8)\text{ }*\text{ 1.30}}=\text{ 1.97}*10^(-4) \end{gathered}`
`\begin{gathered} \text{ pH = }-\text{ log \lparen}\lbrack\text{H}^+\rbrack) \\ \text{ pH = - log \lparen1.97}*10^(-4))\text{ = 3.71 } \end{gathered}`