Question

A 10.0-ml sample of 0.200 m hydrocyanic acid (hcn) is titrated with 0.0998 m naoh. what is the ph at the equivalence point? for hydrocyanic acid, pka = 9.31

Answer 1

When the titration of HCN with NaOH is:

HCN (aq) + OH- (aq) → CN-(aq) + H2O(l)

So we can see that the molar ratio between HCN: OH-: CN- is 1:1 :1

we need to get number of mmol of HCN = molarity * volume 

                      = 0.2 mmol / mL* 10 mL = 2 mmol

so the number of mmol of NaOH = 2 mmol according to the molar ratio

so, the volume of NaOH = moles/molarity

                                          = 2 mmol / 0.0998mL

                                          = 20 mL

and according to the molar ratio so, moles of CN- = 2 mmol

∴the molarity of CN- =  moles / total volume 

                                   = 2 mmol / (10mL + 20mL ) = 0.0662 M

when we have the value of PKa = 9.31 and we need to get Pkb

so, Pkb= 14 - Pka

            = 14 - 9.31 = 4.69 

when Pkb = -㏒Kb

         4.69 = -㏒ Kb 

∴ Kb = 2 x 10^-5

and when the dissociation reaction of CN- is:

CN-(aq) + H2O(l) ↔ HCN(aq) + OH- (aq) 

by using the ICE table:

∴ the initials concentration are:

[CN-] = 0.0662 M

and [HCN] = [OH]- = 0 M

and the equilibrium concentrations are:

[CN-] = (0.0662- X)

[HCN] = [OH-]= X

when Kb expression = [HCN][OH-] /[CN-]

by substitution:

2 x 10^-5 = X^2 / (0.0662 - X)

X = 0.00114 

∴[OH-] = X = 0.00114

when POH = -㏒[OH]

                    = -㏒ 0.00114

POH = 2.94

∴PH = 14 - 2.94 = 11.06