For this problem, we are calculating for the Ka for methylamine with Kb of 4.4x10^{-4}

Recall that Kb and Ka are related via the autoionization constant (1x10^{-14})

$\overline{){\mathbf{Kw}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbf{Kb}}{\mathbf{\times}}{\mathbf{Ka}}}$

Calculate Ka for the conjugate acid of methylamine, CH_{2}NH_{2} (K_{b}= 4.4 x 10^{-4}).

a. 3.91x10^{-11}

b. 6.16x10^{-9}

c. 2.27x10^{-11}

d. 5.51x10^{-9}

e. 1.13x10^{-8}

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