Degree of dissociation of $0.1\,N\,\,C{H_3}COOH$ is (Dissociation constant $ = 1 \times {10^{ - 5}}$)
Degree of dissociation of $0.1\,N\,\,C{H_3}COOH$ is (Dissociation constant $ = 1 \times {10^{ - 5}}$)
- A
${10^{ - 5}}$
- B
${10^{ - 4}}$
- C
${10^{ - 3}}$
- D
${10^{ - 2}}$
Similar Questions
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]
Equal volumes of three acid solutions of $pH \,3, 4$ and $5$ are mixed in a vessel. .........$ \times 10^{-4} \,M$ will be the $H^+$ ion concentration in the mixture ?
Equal volumes of three acid solutions of $pH \,3, 4$ and $5$ are mixed in a vessel. .........$ \times 10^{-4} \,M$ will be the $H^+$ ion concentration in the mixture ?
- [AIPMT 2008]
A monoprotic acid in a $0.1\,\,M$ solution ionizes to $0.001\%$. Its ionisation constant is
A monoprotic acid in a $0.1\,\,M$ solution ionizes to $0.001\%$. Its ionisation constant is
$pH$ 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 $({K_b} = 1.8 \times {10^{ - 5}})$
The ${K_b}$ of ammonia is $1.8 \times {10^{ - 5}}$ at $298$ $K$ temperature. Calculate the $pH$ of $0.1$ $M$ solution.
The ${K_b}$ of ammonia is $1.8 \times {10^{ - 5}}$ at $298$ $K$ temperature. Calculate the $pH$ of $0.1$ $M$ solution.