$pH$ of $0.1\,\, M$ $N{H_3}$ aqueous solution is $({K_b} = 1.8 \times {10^{ - 5}})$
$pH$ of $0.1\,\, M$ $N{H_3}$ aqueous solution is $({K_b} = 1.8 \times {10^{ - 5}})$
- A
$11.13$
- B
$12.5$
- C
$13.42$
- D
$11.55$
Similar Questions
$p{K_a}$ value for acetic acid at the experimental temperature is $5$. The percentage hydrolysis of $0.1\,\,M$ sodium acetate solution will be
$p{K_a}$ value for acetic acid at the experimental temperature is $5$. The percentage hydrolysis of $0.1\,\,M$ sodium acetate solution will be
Calculate $\left[ {{S^{ - 2}}} \right]$ and $\left[ {H{S^{ - 2}}} \right]$ of the solution which contain$0.1$ $M$ ${H_2}S$ and $0.3$ $M$ $HCl$.
[ ${H_2}S$ of ${K_a}\left( 1 \right) = 1.0 \times {10^{ - 7}}$ and ${K_a}\left( 2 \right) = 1.3 \times {10^{ - 13}}$ ]
Calculate $\left[ {{S^{ - 2}}} \right]$ and $\left[ {H{S^{ - 2}}} \right]$ of the solution which contain$0.1$ $M$ ${H_2}S$ and $0.3$ $M$ $HCl$.
[ ${H_2}S$ of ${K_a}\left( 1 \right) = 1.0 \times {10^{ - 7}}$ and ${K_a}\left( 2 \right) = 1.3 \times {10^{ - 13}}$ ]
Calculate $pH$ of solution of $6.0$ $gm$ acetic acid in $250$ $mL$. ( ${K_a} = 1.8 \times {10^{ - 5}}$ at $298$ $K$ ) ( $C = 12, H = 1, O = 16$ )
Calculate $pH$ of solution of $6.0$ $gm$ acetic acid in $250$ $mL$. ( ${K_a} = 1.8 \times {10^{ - 5}}$ at $298$ $K$ ) ( $C = 12, H = 1, O = 16$ )
A weak base $MOH$ of $0.1\, N$ concentration shows a $pH$ value of $9$. What is the percentage degree of ionisation of the base ? ......$\%$
A weak base $MOH$ of $0.1\, N$ concentration shows a $pH$ value of $9$. What is the percentage degree of ionisation of the base ? ......$\%$
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 $0.73 = 5.37$ )
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 $0.73 = 5.37$ )