The $pH$ of $0.01 \ M$ solution of acetic acid is $5.0$. What are the values of $[H^{+}]$ and $K_a$ respectively?
- A$1 \times 10^{-5} \ M, 1 \times 10^{-8}$
- B$1 \times 10^{-5} \ M, 1 \times 10^{-9}$
- C$1 \times 10^{-4} \ M, 1 \times 10^{-8}$
- D$1 \times 10^{-3} \ M, 1 \times 10^{-8}$
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$HClO$ is a weak acid. The concentration of ${H^+}$ ions in $0.1 \ M$ solution of $HClO$ $(K_a = 5 \times 10^{-8})$ will be equal to:
Medium
View SolutionAt $25^{\circ} C$,the percentage of ionization of '$x$' $M$ acetic acid is $4.242$. What is the value of $x$? $(K_a = 1.8 \times 10^{-5})$
The degree of ionization of $0.10 \ M$ lactic acid is $4.0 \%$. The value of $K_a$ is:
What is the $pH$ of $0.001 \,M$ aniline solution? The ionization constant of aniline can be taken from the 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,$(CH_{3})_{2}NH$ $5.4 \times 10^{-4}$ Triethylamine,$(C_{2}H_{5})_{3}N$ $6.45 \times 10^{-5}$ Ammonia,$NH_{3}$ $1.77 \times 10^{-5}$ Quinine $1.10 \times 10^{-6}$ Pyridine,$C_{5}H_{5}N$ $1.77 \times 10^{-9}$ Aniline,$C_{6}H_{5}NH_{2}$ $4.27 \times 10^{-10}$ Urea,$CO(NH_{2})_{2}$ $1.3 \times 10^{-14}$
| Base | $K_{b}$ |
| Dimethylamine,$(CH_{3})_{2}NH$ | $5.4 \times 10^{-4}$ |
| Triethylamine,$(C_{2}H_{5})_{3}N$ | $6.45 \times 10^{-5}$ |
| Ammonia,$NH_{3}$ | $1.77 \times 10^{-5}$ |
| Quinine | $1.10 \times 10^{-6}$ |
| Pyridine,$C_{5}H_{5}N$ | $1.77 \times 10^{-9}$ |
| Aniline,$C_{6}H_{5}NH_{2}$ | $4.27 \times 10^{-10}$ |
| Urea,$CO(NH_{2})_{2}$ | $1.3 \times 10^{-14}$ |
Medium
View SolutionAt $298 \ K$ temperature,calculate the $pH$ of a $0.25 \ M$ solution of $(CH_3)_2NH$ given that its $K_b$ is $5.4 \times 10^{-4}$.
Medium
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