Question

How many liters of 0.547 M koh will be needed to raise the ph of 0.223L of 4.77 M absorbic acid to a ph of 11.11?

Answer 1

M1V1 = M2V2

.0546 V1 = 4.77×.223

V1 = 1.948

POh = - log (OH)

(14= Poh + pH)

POH = 14- 11.11

= 2.89

2.89 = - log (OH)

OH = 7.8 × 10^-2

7.8×10^-2 = 1.065/v ( molarity= no.of mole / volume of solution in liter)

V = 1.065/7.8×10^-2

= 13.66

Total volume required equal to 13.66 + 1.948

=15.60