Question

A 45.00 -mL sample contains 0.754 g of KHC₈H₄O₄, known as KHP . This sample is used to standardize an NaOH solution. At the equivalence point, 43.30 mL of NaOH have been added.

K_a( HC₈H₄O₄^-) = 3.9E-6

a) What was the concentration of the NaOH?

b) What is the pH at the equivalence point?

Answer 1

Answer –

We are given, mass of KHC₈H₄O₄ = 0.754 g , volume = 45.00 mL = 0.045 L

First we need to calculate the moles of KHC₈H₄O₄

Moles of KHC₈H₄O₄ = 0.754 g / 204.22 g.mol^-1

                                   = 0.00369 moles

[KHC₈H₄O₄] = 0.00369 moles / 0.045 L

                      = 0.0820 M

a) At the equivalence point , moles of KHC₈H₄O₄ = moles of NaOH

So, moles of NaOH = 0.00369 moles

Concentration of the [NaOH] = 0.00369 moles / 0.04330 L

                                                     = 0.0853 M

b) pH at the equivalence point

at the equivalence point, moles of KHC₈H₄O₄ = moles of NaOH

so,

KHC₈H₄O₄ (aq) + NaOH (aq) ----> NaKC₈H₄O₄ (aq) + H₂O

0.00369                 0.00369                   0.00369

Moles of NaKC₈H₄O₄ (aq) = moles of KC₈H₄O₄^- (aq) = 0.00369 moles

Total volume = 45 + 43.30 = 88.30 mL

So, [KC₈H₄O₄^- (aq)] = 0.00369 mole / 0.0883 L

                                 = 0.0418 M

Now we need to put ICE table

         KC₈H₄O₄^- (aq) + H₂O --------> KHC₈H₄O₄ (aq) + OH^-

I      0.0418 M                                    0                            0

C        -x                                             +x                         +x

E     0.0418-x                                     +x                        +x

We need to calculate Kb from Ka

We know, Ka * Kb = 1*10^-14

So, Kb = 1*10^-14 / 3.9*10^-6

            = 2.56*10^-9

So, Kb = [KHC₈H₄O₄(aq)] [OH^-] / [KC₈H₄O₄^-(aq)]

2.56*10^-9 = x *x / 0.0418-x

We can neglect x in the 0.0418-x, because Kb value is too small

So,

2.56*10^-9 * 0.0418 = x²

x = 1.035*10^-5 M

x = [OH^-] = 1.035*10^-5 M

so, pOH = -log [OH^-]

              = - log 1.035*10^-5 M

              = 4.98

So, pH = 14-pOH

           = 14 – 4.98

          = 9.02