Question

A 33.6 mL sample of 0.1626 M hydrotelluric acid, H2Te, is titrated with 0.1626 M potassium hydroxide.

What is the pH after 16.8 mL, 33.6 mL, 50.4 mL and 67.2 mL of KOH have been added? The first and second ionization constants for H2Te are 2.3×10-3 and 1.6×10-11

Answer 1

We have 33.6 mL sample of 0.1626M H2Te .So, it has (33.6mL*0.1626M) millimoles of H2Te(millimoles = volume in mL*molarity) = 5.46336 millimoles

If we add 16.8 mL of 0.1626M KOH to this solution we are adding (16.8mL*0.1626M) millimoles of KOH = 2.73168 millimoles

This reaction will occur -

H2Te +KOH -> HKTe + H2O

But as 1 mole of H2Te requires 1 mole of KOH , 2.73168 millimoles of KOH will react with only 2.73168 millimoles of H2Te to produce 2.73168 millimoles of HKTe

Thus the solution obtained will contain 2.73168 millimoles of HKTe and 2.73168 millimoles of H2Te which is an acidic buffer solution.

A buffer solution is a solution of weak acid and its salt with strong base or vice versa. It resists change in pH.

The pH of a buffer solution is given by henderson hasselbalch equation -

pH = pKa + log(salt/acid) where Ka is the equilibrium constant corresponding to the salt which is Ka1 in our case and salt/acid is the ratio of moles of salt and acid which is 2.73168 millimoles/2.73168 millimoles = 1 in this case

so, pH = -log(Ka) + log(1) = -log(2.3*10^-3) = 2.6382

If we add 33.6 mL of 0.1626M KOH to this solution we are adding (33.6mL*0.1626M) millimoles of KOH = 5.46336 millimoles

This reaction will occur -

H2Te +KOH -> HKTe + H2O

But as 1 mole of H2Te requires 1 mole of KOH , 5.46336 millimoles of KOH will react with 5.46336 millimoles of H2Te to produce 5.46336 millimoles of HKTe

Thus the solution obtained will only contain 5.46336 millimoles of HKTe which is an amphiprotic species (it can accept or donate H+ ions )

The pH of a amphiprotic species is given by -

pH = (pKa1 + pKa2)/2 = (-log(2.3*10-3) - log(1.6*10-11))/2 = 13.4341/2 = 6.717

If we add 50.4 mL of 0.1626M KOH to this solution we are adding (50.4 mL*0.1626M) millimoles of KOH = 8.19504 millimoles

This reaction will occur -

H2Te +KOH -> HKTe + H2O

But as 1 mole of H2Te requires 1 mole of KOH , 5.46336 millimoles of H2Te will react with only 5.46336 millimoles of H2Te to produce 5.46336 millimoles of HKTe, remaining KOH = 8.19504 millimoles - 5.46336 millimoles = 2.73168 millimoles of KOH

Now, this HKTe will further react with KOH as -

HKTe + KOH -> K2Te + H2O

But as 1 mole of HKTe requires 1 mole of KOH , 2.73168 millimoles of KOH will react with only 2.73168 millimoles of HKTe to produce 2.73168 millimoles of K2Te

Thus the solution obtained will contain 2.73168 millimoles of K2Te and 2.73168 millimoles of HKTe which is an acidic buffer solution.

A buffer solution is a solution of weak acid and its salt with strong base or vice versa. It resists change in pH.

The pH of a buffer solution is given by henderson hasselbalch equation -

pH = pKa + log(salt/acid) where Ka is the equilibrium constant corresponding to the salt which is Ka2 in our case and salt/acid is the ratio of moles of salt and acid which is 2.73168 millimoles/2.73168 millimoles = 1 in this case

so, pH = -log(Ka2) + log(1) = -log(1.6*10^-11) = 10.7959