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

Answer – We are given, volume = 10.0 mL, [HCN] = 0.200 M

[NaOH] = 0.0998 M

We know the reaction between HCN and NaOH

HCN + NaOH ----> CN^- + H₂O

We need to calculate the moles of HCN = 0.200 M *0.010 L

                                                         = 0.0020 moles

So, HCN + NaOH ----> CN^- + H₂O

     0.0020    0.0020       0.0020 moles

Volume of NaOH needed = 0.0020 moles / 0.0998 M

                                    = 0.02004 L

                                    = 20.04 mL

So total volume = 10+20.04 = 30.04 mL

[CN^-] = 0.0020 moles / 0.03004 L = 0.0665 M

   CN^- + H₂O -----> HCN + OH^-

I 0.0665               0            0

C   -x                  +x          +x

E 0.0665-x          +x           +x

We know, pKa = -log Ka

So, Ka = 10^-9.31

           = 4.90*10^-10

Kb = Kw / Ka

   = 1*10^-14 / 4.90*10^-10

= 2.04*10^-5

We know,

Kb = [HCN] [OH^-] / [CN^-]

2.04*10^-5 = x *x / (0.0665-x)

We can neglect x in the 0.0665-x , since Kb values is too small

So, x² = 2.04*10^-5* 0.0665 M

     x = 1.16*10^-3 M

we know, x = [OH^-] = 1.16*10^-3 M

so, pOH = -log [OH^-]

              = -log 1.16*10^-3 M

              = 2.93

So, pH = 14 – pOH

            = 14 -2.93

           = 11.07