$0.02 \, M$ monobasic acid dissociates $2 \%$,hence,$pH$ of the solution is

For a $10 \ M \ CH_3COOH$ solution,if $K_a = 10^{-5}$,what will be the values of $[H^{+}]$ and $pH$ respectively?

The first and second ionization constants of $H_2X$ are $2.5 \times 10^{-8}$ and $1.0 \times 10^{-13}$ respectively. The concentration of $X^{2-}$ in $0.1 \ M$ $H_2X$ solution is . . . . . . $\times 10^{-15} \ M$. The value of $Y$ is:

Calculate the dissociation constant of a weak monobasic acid if it is $0.05 \%$ dissociated in a $0.02 \ M$ solution.

The ionization constant of $0.1$ $M$ weak acid is $1.74 \times 10^{-5}$ at $298$ $K$ temperature. Calculate the $pH$ of its $0.1$ $M$ solution. (in $.88$)

At $25^{\circ} C$,the percentage of ionization of $x \ M$ acetic acid is $4.242$. What is the pH of the acetic acid solution?
$(\log 4.242=0.6275) ;(\log 0.04242=-1.372) \quad (K_a=1.8 \times 10^{-5})$