what is the ph of a 0.1M acid solution where the acid has a Ka = 10^-5 ?    

1. Let the acid be HX.
2. So,the equation is :

        $$\begin{gathered} c 0 0 \\ HX \to \left[H^{+}\right] + \left[X^{-}\right] \\ c(1-x) x x \\ {K}_a = \frac{\left[H^{+}\right]\left[X^{-}\right]}{\left[HX\right]} \\ {10}^{-5 } =\frac{x^2}{0.1} \\ x = {10}^{-2} \end{gathered}$$
3. [H⁺] = 10-2
 pH = -log[H⁺] = -log[10^-2] = 2