NH4+ ka =5.69x10 HCN ka=6.2 x 10-1 Given the follo be there following the What will...

Question

Question

NH4+ ka =5.69x10 HCN ka=6.2 x 10-1 Given the follo be there following the What will Laathe aqueous approximate pH of an solut

Answer 1

Answer #1

Since NH4CN is salt of weak acid ( HCN ) and weak base ( NH4OH )

Ka ( NH₄⁺ ) = 5.69 X 10^-10

Kb ( CN^- ) = K_w / Ka ( HCN ) = 10^-14 / 6.2 X 10^-10

= 1.6 x 10^-5

since Kb ( CN^- ) > Ka ( HCN ) Solution will be slightly basic.

Answer 2

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