Degree of dissociation of $0.1\,N\,\,C{H_3}COOH$ is (Dissociation constant $ = 1 \times {10^{ - 5}}$)
${10^{ - 5}}$
${10^{ - 4}}$
${10^{ - 3}}$
${10^{ - 2}}$
The $K_a$ of monobasic acid $A, B$ and $C$ are $10^{-6}, 10^{-8}$ and $10^{-10}$ respectively. The concentrations of $A, B$ and $C$ are respectively. $0.1\,M$, $0.01\, M$ and $0.001\, M$. Which of the following is correct for $pOH$ of $A, B$ and $C$ ?
The $pH$ of the solution obtained on neutralisation of $40\, mL\, 0.1\, M\, NaOH$ with $40\, mL\, 0.1\, M\, CH_3COOH$ is
The solution of $N{a_2}C{O_3}$ has $pH$
$K_b$ for $NH_4OH$ is $1.8\times 10^{-5}.$ The $[\mathop O\limits^\Theta H]$ of $0.1\,M\,NH_4OH$ is
Dissociation constant for a monobasic acid is $10^{-4}$ . What is the $pH$ of the monobasic acid ? (If $\%$ dissociation $= 2\,\%$ )