Which solution has the greatest percent ionization?

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$0.10 M {NH}_{3} (K_{b} = 1.8 \times 10^{- 5})$

$0.25 M {HNO}_{2} (K_{a} = 4.5 \times 10^{- 4})$

$1.00 M HCOOH (K_{a} = 1.7 \times 10^{- 4})$

$2.00 M {CH}_{3} {NH}_{2} (K_{b} = 4.4 \times 10^{- 4})$

answer is A.

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Detailed Solution

Ionization constant α to find :

$α = \sqrt{\frac{k_{a}}{c}}$ ; Highest is for b.

$K_{b} = 1.8 \times 10^{- 5} C = 0.1 M$

$α^{2} = \frac{K_{b}}{c} = \frac{1.8 \times 10^{- 5}}{0.1} = 1.8 \times 10^{- 4}$

$\Rightarrow α = 0.0134$

∴ Percentage ionization =1.34% which is highest of the four values given.

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