This question has multiple parts. Work all the parts to get the most points. Cal

Question

This question has multiple parts. Work all the parts to get the most points.

Calculate the pH at the halfway point and at the equivalence point for each of the following titrations.

100.0 mL of 0.30 M  HC7H5O2 (Ka=6.4 X 10^-5) titrated by 0.10 M NaOH

pH at the halfway point =

pH at the equivalence point =

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100.0 mL of 0.30 M C2 H5NH2 (Kb = 5.6 * 10^-4) titrated by 0.40 M HNO3

pH at the halfway point =

pH at the equivalence point =

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100.0 mL of 0.80 M HCl titrated by 0.25 M NaOH

pH at the halfway point =

pH at the equivalence point =

100.0 mL of 0.30 M  HC7H5O2 (Ka=6.4 X 10^-5) titrated by 0.10 M NaOH

pH at the halfway point =

pH at the equivalence point =

Explanation / Answer

a) for the first reaction we solve like this

Ka of acid = 6.4x10-5  and pKa =- log Ka

= - log 6.4x10-5 = 4.1938

At half equivalence point the acid is half neutrlaised to give its conjugate base.

Thus [acid] = [conjugate base] and the solution behaves as an acidic buffer.

The pH of buffer is given by Hendersen equation

pH = pka + log [conjugate base]/[acid]

at half equivalence

pH = pKa = 4.1938

At equivalence

mmoles of acid = mmoles of base

100mL x 0.3M = VmL x 0.1M

thus volume of base = 300mL

and concentration of salt formed = [salt] = 100x0.3 /400= 0.0075 M

the pH of a salt of weak acid and strong base is given by

pH = 1/2[ pKw + pKa + log C]

= 1/2 [14 + 4.1938 + log 0.0075]

= 1/2 [14 + 4.1938 - 2.1249 ] = 8.034

b) in the second part we solve like this

kb of base = 5.6x10-4 and pKb = -log Kb

=-log (5.6x10-4) = 3.25

At half equivalence point the base is half neutrlaised to give its conjugate acid.

Thus [base ] = [conjugate acid ] and the solution behaves as an basic buffer

pOH = pKb and pH = 14 - pOh

pOH = 3.25 and pH = 10.75

At equivalence mmoles of base = mmoles of acid

100mLx 0.3M = V x 0.4 =>V= 75 ml of acid

and volume of acid = 75 mL

and concentraion of salt formed = [salt] = 30/175 = 0.1714 M

and pH of a salt of week base and strong acid is given by

pH = 1/2 [pKw -pKb - log C]

= 1/2 [ 14 -3.25 - log 0.1714]

= 1/2 [ 14 -3.25 + 0.7659 ] = 1/2 [11.5159] = 5.7579

c) in the third reaction we have HCl + NaOH

At half equivalence

1/2 mmoles of acid = mmoles of base used

1/2 x 100mLx0.80 M = V x 0.25M

or volume of NaOh used at half equivalence = 160 mL

The solution has now 40mmol of acid in 260 mL of solution

Thus [H+] = 40/260 = 0.1538 M

pH at equivalence = - log [H+] = -log 0.1538

= 0.8130

As the salt formed is of a strong acid and strong base, the salt is neutral .

pH at equivalence = 7.00

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