Find $pH$ of $5 \times 10^{-3}\, M$ $H_2CO_3$ solution having $10\%$ dissociation
Find $pH$ of $5 \times 10^{-3}\, M$ $H_2CO_3$ solution having $10\%$ dissociation
- A
$3$
- B
$2.7$
- C
$4.3$
- D
$11.3$
Similar Questions
A certain amount of $H_2CO_3$ & $HCl$ are dissolved to form $1$ litre solution. At equilibrium it is found that concentration of $H_2CO_3$ & $CO_3^{-\,-}$ are $0.1\,M$ & $0.01\,M$ respectively. Calculate the $pH$ of solution. Given that for $H_2CO_3$ $K_{a_1} =10^{-5}$ & $K_{a_2} =10^{-8}$
A certain amount of $H_2CO_3$ & $HCl$ are dissolved to form $1$ litre solution. At equilibrium it is found that concentration of $H_2CO_3$ & $CO_3^{-\,-}$ are $0.1\,M$ & $0.01\,M$ respectively. Calculate the $pH$ of solution. Given that for $H_2CO_3$ $K_{a_1} =10^{-5}$ & $K_{a_2} =10^{-8}$
The percentage of pyridine $(C_5H_5N)$ that forms pyridinium ion $(C_5H_5N^+H)$ in a $0.10\, M$ aqueous pyridine solution ($K_b$ for $C_5H_5N = 1.7 \times 10^{-9}$) is
The percentage of pyridine $(C_5H_5N)$ that forms pyridinium ion $(C_5H_5N^+H)$ in a $0.10\, M$ aqueous pyridine solution ($K_b$ for $C_5H_5N = 1.7 \times 10^{-9}$) is
- [NEET 2016]
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 $298$ $K$ ) ( $C = 12, H = 1, O = 16$ )
${K_a} = 1.4 \times {10^{ - 5}}$ of propanoic acid. Calculate its $pH$ of $0.1$ $M$ solution.
${K_a} = 1.4 \times {10^{ - 5}}$ of propanoic acid. Calculate its $pH$ of $0.1$ $M$ solution.
Write characteristic and uses of weak base equilibrium constant ${K_b}$.
Write characteristic and uses of weak base equilibrium constant ${K_b}$.