What is the dissociation constant for NH_4OH | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$\quad\left[\mathrm{H}^{+}ight]=\cdot 10^{-11.27}=10^{-12}, 10^{0.73}=5.37 	imes 10^{-12}$

$\left[\mathrm{OH}^{-}ight]=\frac{\mathrm{K}_{\math

Vedclass · Accepted Answer

$1.75	imes 10^{-5}$

Vedclass · Answer

$\quad\left[\mathrm{H}^{+}ight]=\cdot 10^{-11.27}=10^{-12}, 10^{0.73}=5.37 	imes 10^{-12}$

$\left[\mathrm{OH}^{-}ight]=\frac{\mathrm{K}_{\mathrm{w}}}{\left[\mathrm{H}^{+}ight]}=\frac{7.1 	imes 10^{-15}}{5.37 	imes 10^{-12}}=\frac{7.1}{5.37} 	imes 10^{-3}\, \mathrm{M}=\sqrt{\mathrm{K}_{\mathrm{b}} \mathrm{C}}$

$\left(\frac{7.1}{5.37}ight)^{2} 	imes 10^{-6}=\mathrm{K}_{\mathrm{b}} 	imes 0.1$

$\boxed{{K_b} = 1.747 	imes {{10}^{ - 5}}}$

What is the dissociation constant for $NH_4OH$ if at a given temperature its $0.1\,N$ solution has $pH = 11.27$ and the ionic product of water is $7.1 \times 10^{-15}$ (antilog $0.73 = 5.37$ )

Similar Questions

The ionization constant of benzoic acid is $6.5 \times {10^{ - 5}}$ at $298$ $K$ temperature. Calculate $pH$ of its $0.15$ $M$ solution.

The first and second dissociation constants of an acid $H_2A$ are $1.0 \times 10^{-5}$ and $5.0 \times 10^{-10}$ respectively. The overall dissociation constant of the acid will be

Assuming that the degree of hydrolysis is small, the $pH$ of $0.1\, M$ solution of sodium acetate $(K_a\, = 1.0\times10^{- 5})$ will be

A $0.1\,N $ solution of an acid at room temperature has a degree of ionisation $ 0.1$ . The concentration of $O{H^ - }$ would be

The concentration of $[{H^ + }]$ and concentration of $[O{H^ - }]$ of a $ 0.1$ aqueous solution of $2\%$ ionised weak acid is [Ionic product of water $ = 1 \times {10^{ - 14}}]$