Question

. Strong acid-titrated with strong base. Suppose the titration was reversed in question 2. You titrate 20.0 mL of 0.15 M HCl with 0.15 M NaOH. A. Add this curve to your sketch in question 2 part A. B. Then, determine the pH (a) at the start of the titration and (b) at the equivalence point. (c) What is the total volume of solution at the equivalence point

Answer 1

Answer:

a) attached below

b) i) 0.82  ii) 7

c) 40 mL

Explanation:

A) Titrate 20 mL of 0.15 M HCL with 0.15 M NaOH

sketch attached below

B) Determine PH at

i) start of titration

conc of H^+ = 0.15 M

HCL ------- > H^+  + CL^-

∴ pH = - log ( H^+ )  = - log ( 0.15 ) = 0.82

ii) equivalence point

at equivalence point the moles of acid = moles of base hence pH at equivalence = 7

C) Determine the total volume of solution at the equivalence point

volume of base at equivalence point

= Ma Va = Mb Vb

∴ Vb = ( Ma * Va ) / Mb

        = ( 0.15M * 20 mL ) / 0.15

        = 20.0 mL

Total volume of solution at equivalence point = Ma + Mb = 20 + 20 = 40 mL