Question

Do the calculations for the titration of 50.00 mL of a 0.1000 M solution of H₂SO₃ with a 0.2000 M solution of NaOH. Calculate the pH after the addition of 0.00, 12.50, 25.00, 37.50, 50.00, and 60.00 mL of NaOH. Provide the answers in the blanks below. K_a1(H₂SO₃)=1.23×10^-2; K_a2(HSO₃^-)=6.60×10^-8.

Answer 1

For dissociation of H2SO3,

[H2SO3]= 0.05*0.1 = 0.005

Now, for the first dissociation of H2SO3, Ka1= 1.23*10^-2

Ka1= 1.23*10^-2 = [HSO3-][H+]/[H2SO3]

Say at equilibrium [HSO3-]=[H+]=x,

Solving the value for x, x= 0.0078 M

No, for the second dissociation of HSO3^-, Ka2= 6.6*10^-8

Ka2= 6.6*10^-8 = [SO3^2-][H+]/[HSO3^-]

Say at equilibrium [SO3^2-]=[H+]=x,

Solving the value for x, x= 5.17*10^-10 M

So, the total concentration of H+= 5.17*10^-10 M+ 0.0078 M = 0.0078 M (approx)

Now, for dissociation of NaOH: it is a strong base so 100% dissociation will occur.

After addition of 0.0 ml of NaOH, pH = -log [0.0078] = 2.107

Similarly,

After addition of 12.5 ml of NaOH, moles of OH- = 0.0125*0.2 = 0.0025

So, excess moles of H+ = 0.0053, pH = -log [0.0053] = 2.276

Similarly,

After addition of 25 ml of NaOH, moles of OH- = 0.025*0.2 = 0.005

So, excess moles of H+ = 0.0028, pH = -log [0.0028] = 2.55

Similarly you can calculate the pH for volume of NaOH equal to 37.5, 50.0 and 60.0 ml added to the H2SO3 solution.