Question

25 cm3 of 0.134M benzoic acid solution is titrated with a 0.162M solution of potassium hydroxide at 25 degrees celcius. Ka=6.3x10^-5. Calculate the pH at the following volumes below equivalence point, 0cm3 and 10cm3

Answer 1

Benzoic acid = 25cm3 of 0.134M

number of moles of Benzoic acid = 0.134M x 0.025L = 0.00335 moles

Ka = 6.3x10^-5

a) at equivalent point

KOH = 0.162M

at equivalnet point , volume of KOH = 0.134x25/0.162 = 20.68 cm3

Total volume = 25+20.68 = 45.68cm3 = 0.04568 L

number of moles of KOH = 0..162M x 0.02068L = 0.00335 moles

at equivalent point

number of moles of Benzoic acid is equal to Number of moles of KOH

Ka= 6.3x10^-5

-log(Ka) = -log(6.3x10^-5)

PKa= 4.20

at equivalent point

PH = 7 + 1/2[Pka + logC]

C= number of moles /total volume = 0.00335 / 0.04568 = 0.0733M

PH = 7 + 1/2 [ 4.20 + log(0.0733)]

PH = 8.53

b) at 0 cm3

for weak acids

[H+] = square root of (KaxC)

[H+] = square root of ( 6.3x10^-5 x 0.134)

[H+] = 2.90x10^-3

-log[H+] = - log(2.90x10^-3)

PH = 2.54

c) at 10cm3

KOH = 10cm3 of 0.162M

number of moles of KOH = 0.162M x 0.010L = 0.00162 moles

number of moles of Benzoic acid = 0.134M x 0.025L = 0.00335 moles

                  C6H5COOH   +   KOH   ------------------ C6H5COOK + H2O

                 0.00335             0.00162                           0

              - 0.00162          - 0.00162                       + 0.00162

               0.00173               0                                 + 0.00162

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

PH = 4.20 + log(0.00162/0.00173)

PH = 4.17.