Question

With the titration of 281 mL of 0.501 M carbonic acid (H2CO3) (Ka1 = 4.3 x 10-7, Ka2 = 5.6 x 10-11) with 1.1 M NaOH. How many mL of the 1.1 M NaOH are needed to raise the pH of the carbonic acid solution to a pH of 6.019?

Answer 1

H₂CO₃ + NaOH ---> NaHCO₃ + H₂O             pK_a = 6.37

NaHCO₃ + NaOH ---> Na₂CO₃ + H₂O           pK_a = 10.25

Let y mL NaOH is needed to raise the pH of the carbonic acid solution to a pH of 6.019. At this pH the reaction is governed by the first equation.

Concentration of H₂CO₃ after adding y mL NaOH

= (281*0.501 - y*1.1) / (281+y) = 140.8 -1.1y/(281+y)

Concentration of NaHCO₃ after adding y mL NaOH

= 1.1 y / (281+y)

Mixture of H₂CO₃ and NaHCO₃ will act as a buffer solution. According to Henderson's equation,

pH = pK_a + log [NaHCO₃]/[ H₂CO₃]

or, 6.019 = 6.37 + log (1.1 y)/( 140.8 -1.1y)

or, 1.1y/( 140.8 -1.1y) = 0.445

or, y = 39.5 mL

Addition: When you add NaOH to the H₂CO₃ solution, NaHCO₃ forms according to the equation :

H₂CO₃ + NaOH ---> NaHCO₃ + H₂O             pK_a = 6.37

Now you have a solution of H₂CO₃ and its salt NaHCO₃. This solution acts as a buffer.

For a buffer solution, pH is measured by Henderson's equation:

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

You have the value of final pH = 6.019

pK_a is given = -log[K_a1] = 6.37

You need the concentrations of acid(H₂CO₃) and the salt(NaHCO₃) after addition of y mL of NaOH.

Before the reaction, no of milimoles of acid in the solution = 281 * 0.501 = 140.781 milimoles

No of milimoles of NaOH added = 1.1y milimoles

Total volume of the solution after addition of NaOH = (281 + y) mL

After the reaction,

No of milimoles of acid remained in the solution = 140.781 - 1.1y

Concentration of acid in the final solution = no of milimoles of acid remained/total volume

= [140.781 - 1.1y/(281 + y)] moles/lit

No of milimoles of salt formed after the reaction will be = 1.1y milimoles [as all the NaOH added will convert to salt.

Concentration of salt in the final solution = no of milimoles of salt formed/total volume

= [1.1y/(281 + y)] moles/lit

Now the plug these values in the Henderson equation,

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

or, 6.019 = 6.37 + log [(1.1 y)/ (281 + y)] /[( 140.8 -1.1y)/ (281 + y)]

or, 6.019 = 6.37 + log (1.1 y)/( 140.8 -1.1y)

or, (1.1 y)/( 140.8 -1.1y) = 10^6.019-6.37 = 10^-0.351 = 0.445

or, y = 39.5 mL