Question

A 20.0-mL sample of 0.400 M HBr solution is titrated with 0.400 M NaOH solution. Calculate the pH of the solution after the following volumes of base have been added. (a) 11.6 mL (b) 15.7 mL (c) 20.0 mL (d) 28.2 mL (e) 33.1 mL

Answer 1

HBr + NaOH à NaBr + H₂O

H⁺ + Br^- + Na⁺ + OH^- à Na⁺ +Br^- + H⁺ + OH^-

pH    = - log[HBr]

                = -log[H⁺]

At pH=7,moles of NaOH added = moles of HBr

Initial moles of HBr= Molarity of HBr x Volume of HBr in Litres

During Titration,

Calc. Moles of NaOH added = Molarity of NaOH x Volume of NaOH in Litres

                                        =moles of NaOH added (moles of HBr neutralized)

Moles of HBr left= Initial moles of HBr – Moles of NaOH added

Concn of [HBr]=[H⁺]=moles of HBr left/Total volume of solution

Question(a)

Strength of HBr=0.4M

Volume of HBr taken in titration vessel =20ml =0.02L

Strength of NaOH=0.4M

Before titration, Initial moles of HBr = M_HBr x Vol _HBr

                                                = 0.4 x 0.02

= 0.008 mole HBr

In question A (a), Volume of NaOH added= 11.6ml= 0.0116L

Moles of NaOH added = M_NaOH x Vol _NaOH

                                                = 0.4 x 0.0116

                                = 0.00464 mol = 0.00464 mol HBr

Moles of HBr left = 0.008-0.00464

                        =0.00336 moles

Conc of HBr = moles of HBr left/ total volume of solution

                = 0.00336/(0.02L+0.0116L)

=0.00336/0.0316

=0.1063

pH    = -log[0.1063]

= log[1/0.1063]

= log[9.4073]

=0.97

Therefore pH of HBr solution after adding 11.6ml NaOH is 0.97

In question A (b), Volume of NaOH added= 15.7ml= 0.0157L

Moles of NaOH added = M_NaOH x Vol _NaOH

                                                = 0.4 x 0.0157

                                = 0.00628 mol = 0.00628 mol HBr

Moles of HBr left = 0.008-0.00628

                        =0.00172 moles

Conc of HBr = moles of HBr left/ total volume of solution

                = 0.00172/(0.02L+0.0157L)

=0.00172/0.0357

=0.04818

pH    = -log[0.04818]

= log[1/0.04818]

= log[20.7555]

=1.32

Therefore pH of HBr solution after adding 15.7ml NaOH is 1.32

In question A (c), Volume of NaOH added= 20.0ml= 0.02L

Moles of NaOH added = M_NaOH x Vol _NaOH

                                                = 0.4 x 0.02

                                = 0.008 mol = 0.008 mol HBr

Moles of HBr left = 0.008-0.008

                        =0 moles

This is the endpoint and pH is 7.0 when 20mlof NaOH added.

In question A (d), Volume of NaOH added= 28.2ml= 0.0282L

That is 8.2ml NaOH added after endpoint ,V NaOH=0.0082L

Moles of NaOH = M_NaOH x Vol _NaOH

                                    = 0.4 x 0.0082L

                                = 0.00228 mol OH^-

Conc of excess NaOH = moles of OH^-/ total volume of solution

                = 0.00228/(0.02L+0.0282L)

=0.00228/0.0482

=0.04730

pOH = -log[0.04730]

= log[1/0.04730]

= log[21.1416]

=1.33

As pH + pOH=14,   pH = 14-1.33=12.67

Therefore pH of HBr solution after adding 28.2 ml NaOH is 12.67

In question A (e), Volume of NaOH added= 33.1ml= 0.0331L

That is 13.1 ml of NaOH added after endpoint ,V NaOH=0.0131L

Moles of NaOH = M_NaOH x Vol _NaOH

                                    = 0.4 x 0.0131L

                                = 0.00524mol OH^-

Conc of excess NaOH = moles of OH^-/ total volume of solution

                = 0.00524/(0.02L+0.0331L)

=0.00524/0.0531

=0.09868

pOH = -log[0.09868]

= log[1/0.09868]

= log[10.1338]

=1.01

As pH + pOH=14, pH = 14-1.01=12.99

Therefore pH of HBr solution after adding 33.1 ml NaOH is 12.99