What is the dissociation constant for NH_4OH | vedclass

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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 $0.73 = 5.37$ )

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