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Chemistry Question on Equilibrium

When 0.1000.100 mole of ammonia, NH3NH_3 is dissolved in sufficient water to make 1.0 L1.0 \,L of solution, the solution is found to have a hydroxide ion concentration of 1.34×10−3 M1.34 \times 10^{-3}\, M. Calculate KbK_b for ammonia.

A

1.56×1031.56 \times 10^3

B

1.82×10−51.82 \times 10^{-5}

C

1.37×10−31.37 \times 10^{-3}

D

1.28×1051.28 \times 10^5

Answer

1.82×10−51.82 \times 10^{-5}

Explanation

Solution

NH3+H2O⇌NH4++OH−NH_{3}+H_{2}O { \rightleftharpoons} NH_{4}^{+}+OH^{-} At equi. (0.100−1.34×10−3)M1.34×10−3 M1.34×10−3 M=0.09866 M(0.100-1.34 \times 10^{-3})M 1.34\times 10^{-3}\,M 1.34 \times 10^{-3}\,M = 0.09866\, M Kb=[NH4+[OH−]][NH3]=1.34×10−3×1.34×10−30.09866K_{b}=\frac{\left[NH_{4}^{+}\left[OH^{-}\right]\right]}{\left[NH_{3}\right]}=\frac{1.34\times10^{-3}\times1.34\times10^{-3}}{0.09866} =1.8199×10−5≈1.82×10−5=1.8199 \times 10^{-5}\approx1.82 \times10^{-5}