What is $[N H_{4}^{+}]$ $[NH_4^+]$ in a solution that is 0.0200 M $N H_{3}$ $NH_3$ and .0100 M $K O H$ $KOH$ ?

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Answer 1 · 2016-06-20T17:00:25

$[N H_{4}^{+}] = 3.54 \times 10^{- 5} mol/l$ $[NH_4^+]=3.54xx10^(-5)" ""mol/l"$

Ammonia is a weak base and partially ionises in water:

$N H_{3} + H_{2} O ⇌ N H_{4}^{+} + O H^{-}$ $NH_3+H_2OrightleftharpoonsNH_4^++OH^-$

For which:

$K_{b} = \frac{[N H_{4}^{+}] [O H^{-}]}{[N H_{3}]} = 1.77 \times 10^{- 5} mol/l$ $K_b=([NH_4^+][OH^-])/([NH_3])=1.77xx10^(-5)" ""mol/l"$

$:.[NH_4^+]=K_bxx[[NH_3]]/[[OH^-]]$

The concentrations given are at equilibrium so:

$[NH_4^+]=1.77xx10^(-5)xx0.02/0.01=3.54xx10^(-5)" ""mol/l"$

What is [NH_4^+][NH+4][NH_4^+] in a solution that is 0.0200 M NH_3NH3NH_3 and .0100 M KOHKOHKOH?