Question

Consider the titration of a 34.0mL sample of 0.180M  HBr with 0.210M KOH. Determine each of the following:

Part A

the initial pH

Part B

the volume of added base required to reach the equivalence point

Express your answer in milliliters.

Answer 1

Part-A :-

At initial stage there is only HBr in the solution so the pH of the solution is affected by the concentration of H⁺ in HBr.

The concentration of HBr = 0.180 M

              HBr H⁺ + Br ^-

1 mole of HBr produces 1 mole of H⁺

So concentration of H⁺ = concentration of HBr = 0.180 M

             [H⁺] = 0.180 M

We know pH = - log [H⁺]

                    = - log 0.180
                    = 0.74

Therefore the initial pH of the solution is 0.74

Part- B :-

Number of moles of HBr , n = Molarity x volume in L

                                            = 0.180 M x 34.0x10 ^-3 L                  Since 1 mL = 10 ^-3 L

                                            = 6.12x10^-3 mol

HBr + KOH KBr + H2O

According to the balanced equation ,

1 mole of HBr reacts with 1 mole of KOH

6.12x10^-3 mol of HBr reacts with 6.12x10^-3 mol of KOH

So volume of base required to reach the equivalence point is ,

V = number of moles of base / Molarity of base

    = 6.12x10^-3 mol / 0.210 M

   = 0.0291 L

   = 0.0291 x 1000 mL

   = 29.1 mL

Therefore the volume of added base required to reach the equivalence point is 29.1 mL