The $pH$ of $0.005 \,M$ codeine $\left( C _{18} H _{21} NO _{3}\right)$ solution is $9.95 .$ Calculate its ionization constant and $p K_{ b }$
The $pH$ of $0.005 \,M$ codeine $\left( C _{18} H _{21} NO _{3}\right)$ solution is $9.95 .$ Calculate its ionization constant and $p K_{ b }$
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$0.01\, M \,HA(aq.)$ is $2\%$ ionized, $[OH^-]$ of solution is :-
$0.01\, M \,HA(aq.)$ is $2\%$ ionized, $[OH^-]$ of solution is :-
When $CO_2$ dissolves in water, the following equilibrium is established
$C{O_2} + 2{H_2}O\, \rightleftharpoons {H_3}{O^ + } + HCO_3^ - $
for which the equilibrium constant is $3.8 \times 10^{-7}$ and $pH = 6.0$. The ratio of $[HCO_3^- ]$ to $[CO_2]$ would be :-
When $CO_2$ dissolves in water, the following equilibrium is established
$C{O_2} + 2{H_2}O\, \rightleftharpoons {H_3}{O^ + } + HCO_3^ - $
for which the equilibrium constant is $3.8 \times 10^{-7}$ and $pH = 6.0$. The ratio of $[HCO_3^- ]$ to $[CO_2]$ would be :-
What is the $ pH$ of $0.01\, M$ glycine solution? For glycine, $K{a_1} = 4.5 \times {10^{ - 3}}$ and $K{a_2} = 1.7 \times {10^{ - 10}}$ at $298 \,K$
What is the $ pH$ of $0.01\, M$ glycine solution? For glycine, $K{a_1} = 4.5 \times {10^{ - 3}}$ and $K{a_2} = 1.7 \times {10^{ - 10}}$ at $298 \,K$
- [AIIMS 2004]
Given the two concentration of $HCN (K_a = 10^{-9})$ are $0.1\,M$ and $0.001\,M$ respectively. What will be the ratio of degree of dissociation ?
Given the two concentration of $HCN (K_a = 10^{-9})$ are $0.1\,M$ and $0.001\,M$ respectively. What will be the ratio of degree of dissociation ?
The solubility of a salt of weak acid $( A B )$ at $pH 3$ is $Y \times 10^{-3} mol L ^{-1}$. The value of $Y$ is
. . . . . (Given that the value of solubility product of $A B \left( K _{ sp }\right)=2 \times 10^{-10}$ and the value of ionization constant of $H B \left( K _{ a }\right)=1 \times 10^{-8}$ )
The solubility of a salt of weak acid $( A B )$ at $pH 3$ is $Y \times 10^{-3} mol L ^{-1}$. The value of $Y$ is
. . . . . (Given that the value of solubility product of $A B \left( K _{ sp }\right)=2 \times 10^{-10}$ and the value of ionization constant of $H B \left( K _{ a }\right)=1 \times 10^{-8}$ )
- [IIT 2018]