Question

Calculate the pH in the titration of 25.0 mL of 0.100M HCl with 0.200M NaOH. Finally, calculate the pH after 18.75 mL of base have been added.

Answer 1

NaOH + HCl -------> NaCl + H2O

1) Initial pH

[ HCl ] = 0.100M

[ H+ ] = 0.100M

pH = -log[H+]

= - log (0.100)

= 1

2) pH at equivalencepoint

NaOH is strong base

HCl is strong acid

so, at equivalence point, pH = 7

3) pH after addition of 18.75ml of NaOH

No of mole of HCl =( 0.100mol/1000ml)×25ml =0.0025

According to stoichiometry 0.0025mol of HCl require 0.0025mol of NaOH

Volume of NaOH solution having 0.0025mol of NaOH = (1000ml/0.200)×0.0025=12.5ml

So, the equivalence point = 12.5ml

Volume of NaOH addition after equivalence point =18.75ml - 12.5ml = 6.25ml

No of mole of NaOH in 6.25ml = (0.200mol/1000ml)×6.25ml = 0.00125

Total volume = 25ml + 18.75ml = 43.75ml

[OH- ] = (0.00125mol/43.75ml)×1000ml = 0.0286M

pOH = - log [ OH- ]

= - log ( 0.0286)

= 1.54

pH + pOH = 14

Therefore,

pH = 14- 1.54

= 12.46