$0.01 \, M$ acetic acid solution is $1 \%$ ionised,then $pH$ of this acetic acid solution is :

The ionization constant of benzoic acid is $6.5 \times 10^{-5}$ at $298 \ K$ temperature. Calculate $pH$ of its $0.15 \ M$ solution. (in $.5$)

The $pH$ of $0.05 \ M$ acetic acid is $(K_a = 2 \times 10^{-5})$.

The percent dissociation of a weak monobasic acid is $3 \%$ in its $0.02 \ M$ solution. What is the dissociation constant of the acid?

The $pH$ of $0.1 \ M$ acetic acid is $3$. The dissociation constant of the acid will be:

Sulphurous acid $(H_{2}SO_{3})$ has $Ka_{1} = 1.7 \times 10^{-2}$ and $Ka_{2} = 6.4 \times 10^{-8}$. The $pH$ of $0.588 \ M \ H_{2}SO_{3}$ is ..... . (Round off to the Nearest Integer)