The degree of dissociation of $0.1\,M\,HCN$ solution is $0.01\%$ . Its ionisation constant would be

  • A

    ${10^{ - 3}}$

  • B

    ${10^{ - 5}}$

  • C

    ${10^{ - 7}}$

  • D

    ${10^{ - 9}}$

Similar Questions

${K_{C{H_3}COOH}} = 1.9 \times {10^{ - 5}}$. Calculate $pH$ at end point in titration of $0.1$ $M$ $C{H_3}COOH$ and $0.1$ $M$ $NaOH$.

Values of dissociation constant, $K_a$ are given as follows

      Acid       $K_a$
      $HCN$       $6.2\times 10^{-10}$
      $HF$       $7.2\times 10^{-4}$
      $HNO_2$       $4.0\times 10^{-4}$

Correct order of increasing base strength of the base $CN^-,F^-$ and $NO_2^-$ will be

  • [JEE MAIN 2013]

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 $C$ ?

What is the $pH$ of $0.001 \,M$ aniline solution? The ionization constant of aniline can be taken from Table . Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.

Base $K _{ b }$
Dimethylamine, $\left( CH _{3}\right)_{2} NH$ $5.4 \times 10^{-4}$
Triethylamine, $\left( C _{2} H _{5}\right)_{3} N$ $6.45 \times 10^{-5}$
Ammonia, $NH _{3}$ or $NH _{4} OH$ $1.77 \times 10^{-5}$
Quinine, ( $A$ plant product) $1.10 \times 10^{-6}$
Pyridine, $C _{5} H _{5} N$ $1.77 \times 10^{-9}$
Aniline, $C _{6} H _{5} NH _{2}$ $4.27 \times 10^{-10}$
Urea, $CO \left( NH _{2}\right)_{2}$ $1.3 \times 10^{-14}$

The dissociation constants of two acids $HA_1$ and $HA_2$ are $3.0 \times 10^{-4}$ and $1.8 \times 10^{-5}$ respectively. The relative strengths of the acids will be