Question

What is the pH at the equivalence point when 25.00 mL of a 0.150 M solution of acetic acid (CH3COOH) is titrated with 0.10 M NaOH to its end point?

Answer 1

Answer – We are given, [CH₃COOH] = 0.150 M , volume = 25.00 mL , [NaOH] = 0.10 M

Now we need to calculate the moles of CH₃COOH-

moles of CH₃COOH = 0.150 M * 0.025 L

                                   = 0.00375 moles

At equivalence point, moles of acid = moles of base

So, moles of NaOH = 0.00375 moles

So, volume of the NaOH used is –

Volume of NaOH = 0.00375 moles / 0.10 M

                              = 0.0375 L

                               = 37.5 mL

When we added NaOH up to equivalence point, then all acid converted to CH₃COO^- and there is formed conjugate base of CH₃COOH.

Reaction – CH₃COOH + NaOH -----> CH₃COO^- + H₂O

                  0.00375        0.00375            0.00375

Moles of conjugate base form, CH₃COO^- = 0.00375 moles

Total volume = 25+37.5

                      = 62.5 mL

[CH₃COO^-] = 0.00375 moles / 0.0625 L

                   = 0.060 M

Now we need to put ICE chart for this one –

CH₃COO^- + H₂O ------> CH₃COOH + OH^-

I   0.06                               0              0

C    -x                               +x             +x

E 0.06-x                            +x            +x

W need to calculate Kb value from given Ka value

We know, Kb =1*10^-14 / 1.8*10^-5

                        = 5.55*10^-10

So, Kb = [CH₃COOH] [OH^-] / [CH₃COO^-]

5.55*10^-10 = x *x / (0.06-x)

We can neglect x in the 0.06-x, since 5 % rules and Ka value also too small

So, x² = 5.55*10^-10*0.06

        x = 5.77*10^-6

so, [OH^-] = x = 5.77*10^-6

so pOH = -log [OH^-]

             = - log 5.77*10^-6

               = 5.24

So, pH = 14-pOH

            = 14-5.24

            = 8.76

So, the pH at the equivalence point when 25.00 mL of a 0.150 M solution of acetic acid (CH3COOH) titrated with 0.10 M NaOH to its end point is 8.76.