Question

Using ICE tables and acid-base equations, compare the change in pH that occurs when 10 mL of a 0.10 M NaOH is added to 100 mL of pure water, and when it is added to 100 mL of a benzoic acid buffer soltution that is 0.15 M in both C₆H₅CO₂H and C₆H₅CO₂Na. *Hint: M1V1=M2V2

Answer 1

a) When NaOH is added to pure water then it means that NaOH solutio is being diluted to 110mL.

We will use millmole equivalene before and after dilution to calculate the new concentration of NaOH

M1V1 = M2V2

M1 = 0.1M

V1 = 10mL

M2 = ?

V2 = 110mL

Putting values

0.1 X 10 = M2 X 110

M2 = new concentration = 0.0091 M

Hence [OH-] = [NaOH] = 0.0091

pOH = -log [OH-] = -log[0.0091] = 2.04

pH = 14- pOH = 14-2.04 = 11.96

b) When NaOH is added to acidic buffer then it reacts with the acid of the buffer to form more salt

here the moles of NaOH added = molarity X volume = 0.1 x 10 = 1millimoles

so 1millimoles of NaOH will react with 1millimoles of benzoic acid to form new 1millimoles of sodium benzoate

NaOH + C₆H₅CO₂H ---> C₆H₅CO₂Na + H2O

initial moles of C₆H₅CO₂H = molarity X volume = 0.15 x 100 = 15 millimoles

initial moles of C₆H₅CO₂Na = molarity X volume = 0.15 X 100 = 15 millimoles

After addition of NaOH

moles of C₆H₅CO₂H = 15 millimoles - 1millimole = 14 millimoles

moles of C₆H₅CO₂Na = 15 millimoles + 1millimole = 16 millimoles

pKa of benzoic acid = 4.20

we will use Henersen Hassalbalch's equation for buffer to calculate pH

pH = pKa + log [salt] / [acid]

pH = 4.20 + log [16]/[14]

pH = 4.20 + 0.058 = 4.258