Question

Find the pH during the titration of 20.00 mL of 0.2850 M benzoic acid, C₆H₅COOH (K_a = 6.3  10^-5), with 0.2850 M NaOH solution after the following additions of titrant.

(a) 0 mL

(b) 10.00 mL

(c) 15.00 mL

(d) 20.00 mL

(e) 25.00 mL

Answer 1

Answer.

a)

Millimoles of benzoic acid = 20 x 0.285 = 5.7

pKa = 4.2

after adding 0 mL :

[H+] = sqrt (Ka x C)

= sqrt ( 6.3 x 10^-5 x 0.285)

= 4.23 x 10^-3 M

pH = -log [H+] = -log [4.23x 10^-3 ]

= 2.37

pH = 2.37

b) after adding 10.00 ml NaOH :

millimoles of NaOH = 10 x 0.285 = 2.85

C6H5COOH + NaOH ---------------> CH3COONa + H2O

3 2.85 0 0

2.85 0 2.85 2.85

it is half- equivalence point. so pH

pH = pKa

pH = 4.20

c) after adding 15 ml NaOH :

millimoles of NaOH = 15 x 0.285 = 4.275

C6H5COOH + NaOH ---------------> CH3COONa + H2O

3 4.275 0 0

0.75 0 4.275 4.275

pH = pKa + log[salt/acid]

pH = 4.20 + log[4.275/0.75]

pH = 4.96

a) after adding 20.00 mL :

NaOH millimoles = 20 x 0.285 = 5.7

it is half equivalence point . so here salt only remains. so

salt concentration = 5.7 / 20 + 20 = 0.1425 M

pH = 7 + 1/2 (pKa + log C)

= 7 + 1/2 (4.2 + log 0.1425)

pH = 8.58

e)

25 ml NaOH :

millimoles = 25 x 0.285 = 7.125

here strong base reamined. so

[OH-] = 7.125 - 5.7 / (20 + 25) = 0.032 M

pOH = -log [OH-] = -log [0.032] = 1.49

pH =10.43.

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