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 .

5.94

Explanation:

At equilibrium

⇒[CH3NH^+] = 0.130/2= 0.065 M

hence we can write that

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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