Question

Given: A 0.147 gram sample of an unknown weak monoprotic organic acid, HA, was dissolved in sufficient water to make 50.0 milliliters of solution and was titrated with a 0.0363 M NaOH solution. After the addition of 10.6 mL of base, a pH of 5.65 was recorded. The equivalence point was reached after the addition of 42.2 mL total of the 0.0363 M NaOH.

1. Calculate the number of moles of acid in the original sample.
2. Calculate the molecular weight of the acid HA.
3. Calculate the value of the ionization constant (Ka) of the acid HA
4. Calculate the pH before any base was added
5. Calculate the pH after 57.7 mL total of base has been added
6. Calculate the pH at the equivalence point.

Answer 1

a)moles of acid = moles of base used for equivalence

= 42.2 x10^-3 x0.0363

= 1.531 x10^-3

b) moles of acid = mass/ molar mass

1.531 x10^-3 = 0.147/M

molar mass = 95.96 g/mol

molarity of acid = moles /V(L)

= 1.531x10^-3 x1000/50

=0.0306 M

c) HA + OH- -------------> A- + H2O

50x0.0306 -- 0 ---- initial mmoles

= 1.531   

10.6 x 0.0363 =0.385 change

1.15 0 0.385 - equilibriummoles

The solution is a buffer , whose pH is calculated using Hendersen equation

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

5.65 = pKa + log 0.385/1.15

Hnec pKa = 6.125 and Ka = 7.5x10^-7

4) P=pH of the acid before the base was added

pH = 1/2 pKa -1/2 log C

= 1/2 (6.125) - 1/2 log(0.0306)

= 3.8196

5) pH after 57.7 mL base added

  HA + OH- -------------> A- + H2O

50x0.0306 -- 0 ---- initial mmoles

= 1.531   

57.7x 0.0363 = 2.087 change

0 0.556 1.531 - equilibriummoles

thus the solution has a strong base and a salt

thus pH is calculated fron [OH-] from base

[OH-] from base = 0.556 mmol/ 107.7mL

=0.005164 M

Thus the pH = 14 -pOH

and pOH = 2.2869

and pH = 14 - 2.2869

= 11.713

6) pH at equivalence point

  HA + OH- -------------> A- + H2O

50x0.0306 -- 0 ---- initial mmoles

= 1.531   

42.2 x0.0363 change

0 0 1.531 0 equilibrium mmol

the solution contains only salt of weak ans strong base , which is basic in nature.

[Cojugate base] = [salt] = 1.531 /92.2 = 0.0166 M

the pH of the salt is given by

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

= 1/2 [ 14+ 6.125 + log 0.0166]

= 9.1726