Evaluate the pH of 0.05M NaCN. When the kb =10 ^-4.7?

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Answer 1 · 2018-05-04T10:37:35

Given,

#[CN^-] = 0.05"M"#

#K_"b" = 1*10^(-4.7)#

Let's consider the equilibrium corresponding to the preceding data,

#CN^(-)(aq) rightleftharpoons OH^(-)(aq) + HCN(aq)#

And calculate the approximate #[OH^-]#,

#K_"b" = ([OH^-][HCN])/([CN^-]) = 1*10^(-4.7)#

#=>[CN^-]*K_"b" = x^2#

#=> x = [HCN] = [OH^-] approx 1.0*10^-3M#

Now, recall,

#K_"w" = [OH^-][H^+] = 1*10^-14#

Hence,

#"pH" = -log[H^+] = -log(K_"w"/([OH^-])) approx 11.0#