What is the dissociation constant for $NH_4OH$ if at a given temperature its $0.1\,N$ solution has $pH = 11.27$ and the ionic product of water is $7.1 \times 10^{-15}$ (antilog $0.73 = 5.37$ )

The $pH$ of $ 0.1 \,M$  solution of a weak monoprotic acid  $1\%$  ionized is

The first ionization constant of $H _{2} S$ is $9.1 \times 10^{-8}$. Calculate the concentration of $HS ^{-}$ ion in its $0.1 \,M$ solution. How will this concentration be affected if the solution is $0.1\, M$ in $HCl$ also? If the second dissociation constant of $H _{2} S$ is $1.2 \times 10^{-13}$, calculate the concentration of $S^{2-}$ under both conditions.

The $ pH$  of $ 0.1$ $M$ acetic acid is $3$, the dissociation constant of acid will be

The ${K_b}$ of ammonia is $1.8 \times {10^{ - 5}}$ at $298$ $K$ temperature. Calculate the $pH$ of $0.1$ $M$ solution. 

The concentration of $[{H^ + }]$ and concentration of $[O{H^ - }]$ of a $ 0.1$ aqueous solution of $2\%$ ionised weak acid is [Ionic product of water $ = 1 \times {10^{ - 14}}]$