Question

What is the pH when 2.77 moles of HN3 and 4.45 moles of N3- are mixed in 102 mL H2O? The Ka of HN3 is 1.9 x 10^-5.

Answer 1

Ka of HN3 = 1.9*10^-5

PKa = -logKa

        = -log1.9*10^-5

          = 4.7212

[HN3]   = no of moles/volume in L   = 2.77/0.102

[N3^-]   = no of moles/volume in L   = 4.45/0.102

PH = Pka + log[HN3]/[N3^-]

        = 4.7212 + log(2.77/0.102)/(4.45/0.102)

       = 4.7212+ log2.77/4.45

       = 4.7212 -0.2058

      = 4.5154>>>>answer