At 298 K the base dissociation constants in water are K_b(CH3NH2) = 4.4 x 10^-4 and K_b(C6H5NH2) = 4.2 x 10^-10. Calculate pK_b(C6H5NH2) - pK_b(CH3NH2). Give the answer correct to two decimal places.

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JEE Main levelNumerical answerDifficulty 2 of 3nc120

At 298 K the base dissociation constants in water are $K_{b} (CH X_{3} NH X_{2}) = 4.4 \times 1 0^{- 4}$ and $K_{b} (C X_{6} H X_{5} NH X_{2}) = 4.2 \times 1 0^{- 10}$ . Calculate $p K_{b} (C X_{6} H X_{5} NH X_{2}) - p K_{b} (CH X_{3} NH X_{2})$ . Give the answer correct to two decimal places.

Numerical answer type, tolerance 0.02.

Answer: 6.02

pKb = -log Kb. For methylamine pKb = 3.357 and for aniline pKb = 9.377. The difference is 6.020, i.e. 6.02, which equals log10(4.4e-4 / 4.2e-10), showing that aniline is about a million times less basic.

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At 298 K the base dissociation constants in water are Kb​(CHX3​NHX2​)=4.4×10−4 and Kb​(CX6​HX5​NHX2​)=4.2×10−10. Calculate pKb​(CX6​HX5​NHX2​)−pKb​(CHX3​NHX2​). Give the answer correct to two decimal places.

Answer: 6.02

More questions on Basicity of amines

At 298 K the base dissociation constants in water are $K_{b} (CH X_{3} NH X_{2}) = 4.4 \times 1 0^{- 4}$ and $K_{b} (C X_{6} H X_{5} NH X_{2}) = 4.2 \times 1 0^{- 10}$ . Calculate $p K_{b} (C X_{6} H X_{5} NH X_{2}) - p K_{b} (CH X_{3} NH X_{2})$ . Give the answer correct to two decimal places.