Question

Consider HCN (Ka= 5.8 x 10^-10) and NaCN

a. What is the pH of a buffer obtained by adding 9.8grams of NaCN (MW=49 g/mole) to 100ml of .20 M HCN?

b. What is the pH of this buffer after .01 moles of OH- are added to this solution?

Answer 1

no of moles of NaCN   = W/G.M.Wt

                            = 9.8/49   = 0.2moles

no od moles of HCN   = molarity * volume in L

                               = 0.2*0.1 = 0.02 moles

PKa   = -logKa

          = -log5.8*10^-10   = 9.2365

PH     = PKa +log[NaCN]/HCN]

         = 9.2365 +log0.2/0.02

         = 9.2365 +1

        = 10.2365

no of moles of NaCN after addition 0.01moles of OH^-   = 0.2+0.01   = 0.21moles

no of moles of HCn after addition of 0.01 moles of OH^- = 0.02-0.01 = 0.01moles

PH    = PKa + log[NaCN]/[HCN]

       = 9.2365+log0.21/0.01

      = 9.2365+1.3222   =10.5587