A decimolar solution of ammonium hydroxide is ionised to the extent of 1.3%. If $\log{1.3}$. If $\log{1.3 = 0.11}$, what is the pH of the solution?

Question

Dr. Abubacker · Accepted Answer

(a)
$NH_{4}OH ightleftharpoons NH_{4}^{+} + OH^{-}$
At $t = 00.1\ \ \ \ \ \ M\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0\ \ \ \ \ \ \ \ \ \ 0$
At equilibrium $\left( 0.1 - \frac{1.3 	imes 0.1}{100}M ight)\ \ 0.0013\ M(0.0013\ M)$
$= 0.1\$M
$= \left\lbrack OH^{-} ightbrack = 0.0013\ M = 13 	imes 10^{- 4}$ M
$pOH = - \log{\left\lbrack OH^{-} ightbrack	ext{=} - \log{\left( 13 	imes 10^{- 4} ight) = 2.89}}$
$pH = 14 - 2.89 = 11.11$

Q. A decimolar solution of ammonium hydroxide is ionised to the extent of 1.3%. If $\log{1.3}$. If $\log{1.3 = 0.11}$, what is the pH of the solution?

Explanation

Reviewed by Dr. Abubacker

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