Question

CO3(2-) is dibasic with pKb1 = 3.67 and pKb2 =7.65.

75.00 mL of 0.1130 M Sodium Carbonate, Na2CO3, was titrated with 0.2260 M HCl. Calculate the pH of the solution when 10.00, 16.00, 37.50, 82.50 mL of HCl added. Please explain each step. Thank you!

Answer 1

mmol of Na2CO3 = MV = 75*0.113 = 8.475

a)

mmol of acid = MV = 10*0.2260 = 2.26

after reaction

mmol of CO3-2 = 8.475-2.26 = 6.215

mmol of HCO3- formed = 2.26

pH = pKa2 + log(CO3-2/HCO3-)

pH = 10.36 + log(6.215/2.26) = 10.81

b)

mmol of acid = MV = 16*0.2260 = 3.616

after reaction

mmol of CO3-2 = 8.475-3.616= 4.859

mmol of HCO3- formed = 3.616

pH = pKa2 + log(CO3-2/HCO3-)

pH = 10.36 + log(4.859/3.616) = 10.488

c)

mmol of acid = MV = 37.5*0.2260 = 8.475

after reaction

mmol of CO3-2 = 8.475-8.475= 8.475

mmol of HCO3- formed = 8.475

pH is obtained via

pH = 1/2*(pKA1 + pKA2)

pH = 1/2*(6.37 + 10.33) = 8.35

d)

mmol of acid = 82.5*0.2260 = 18.645

mmol of acid left = HCl - Na2CO3 = 18.645 - 2*8.475 = 1.695

Vtotal = 82.5+75

[H+] = 1.695 / ( 82.5+75) = 0.010761

ph = -log(0.010761) = 1.968