13 1000 ml of 0200 M NaOH was titrated with 0200 M H2SO4 Cal

13. 100.0 ml of 0.200 M NaOH was titrated with 0.200 M H2SO4. Calculate the pH before the reaction starts, after addition 50 ml of acid, and after addition of 75.0 ml of acid.

Solution

1) before reaction starts :

[ OH- ] = 0.200 M

pOH = -log [OH-] = -log (0.200)

pOH = 0.7

pH + pOH = 14

pH = 13.30

2) 50 mL acid added

millimoles of OH - = 100 x 0.2 = 20

millimoles of H+ = 0.2 x 2 x 50 = 20

both millimoles are equal so here pH = 7.00

3) 75 mL acid added

millimole sof H+ = 2 x 75 x 0.2 = 30

[H+] left = 30 -20 / 75 + 100 = 0.057 M

pH = -log (0.057)

pH = 1.24