Question

What will the pH of a solution be when 0.9701 L of NaOH (pH = 14.409) is added to 0.357 L of 2.89 M ascorbic acid (H₂C₆H₆O₆)?

Answer 1

PH of NaO H= 14.409

POH+PH= 14

POH= 14-PH=14-409

POH= -0.409

-log[OH-]= -0.409

[OH-]= 10^-0.409

[OH-]= 0.3899M

[OH-] 0.390M

volume of NaOH= 0.9701 L

number of moles o f NaOH= 0.390M x0.9701L=0.378 mole

number of moles of NaOH= 0.378 moles

ascoribic acdi = 0.357L of 2.89 L]

number of moles of ascorbic acid= 2.89Mx0.357L

number of moles of ascorbic acid= 1.032 moles\

C6H8O6 + NaOH------------ C6H5O6Na + H2O

1.032          0.378                   0

-0.378        -0.378                + 0.378

0.654              0                 + 0.378

after addition of NaOH ,

number of moles of ascorbic acid= 0.654 moles

number of moles of salt = 0.378moles

pka of ascorbic acid= 4,17

PH= PKa + log[salt]/[acid]

PH= 4.17 + log(0.378/0.654)

PH= 3.93.