Question

50 ml sample of 0.0950 M acetic acid (ka: 1.8* 10^{-5) isbeing titrated with 0.106 M NaoH.}

1- what is the PH at the midpoint of titration?

2- what is the PH equivalence point of the titration?

3-what is the PH and endpoint of titration/

Answer 1

Given data

ka = 1.8* 10^-5

    pKa =4.74        

Volume of acetic acid, V1 = 50 mL

    Molarity of acetic acid M1 = 0.095M

Molarity of NaOH, M2 = 0.106 M

∴ The volume of NaOH are required to reachthe equivalence point, V2 = M1V1 / M2

                                                                                                                = 0.095 M * 50 mL / 0.106 M

                                                                                                                 =44.81 mL

At mid point of titartion number of moles of aceticacid = number of moles of sodium acetate

   ∴ pH = pKa + log[salt] /[acid]

[salt] /[acid] = 1 so  log[salt] /[acid] =0

    Hence pH = pKa

                    = 4.74

b) At equivalence point of the titration all the acidis converted into salt

∴ Concentration of salt = molarity of acid *volume of acid / total volume

                                      =0.095 M * 50 mL / ( 50 mL + 44.81 mL )

                                  C= 0.05 M

   Sodium acetate ca acts as a base

    The Kb of acetate ion = Kw/ Ka

                                     = 1.0*10^-14 / 1.8*10^-5

                                     = 5.56*10^-10

The concentration of OH- ion at equivalence point =√Kb*C

                                                                              =(5.56*10^-10 * 0.05)^1/2

                                                                             = 5.27*10^-6 M

                        ∴ pOH = - log [OH^-]

                                      =- log( 5.27*10^-6 M)

                                     = 5.278

                                pH = 14 - 5.278

                                      = 8.722

c) Similar to b)