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Calculate the pH at the equivalence point for the titration of 0.130 M methylamine (CH3NH2) with 0.130 M HCl. The K b Kb of methylamine is 5.0 × 10 − 4 .

Respuesta :

Manetho

Answer:

5.94

Explanation:

At equilibrium

[tex]CH_3NH_3^+\rightarrow CH_3NH_2+ H^+[/tex]

[tex]K_a=\frac{K_w}{K_b} = \frac{[CH_3NH_2][H^+] }{CH_3NH_3^+}[/tex]

⇒[CH3NH^+] = 0.130/2= 0.065 M

hence we can write that

[tex]\frac{10^{-14}}{5\times10^{-4}} = \frac{x^2}{0.065}[/tex]

x= concentration of H+ ions

⇒2×10^(-11)×0.065 = x^2

⇒x= 1.14×10^{-6}

therefore ph=log [H+]= log[ 1.14×10^{-6}]= 5.94

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