Given: A 0.147 gram sample of an unknown weak monoprotic organic acid, HA, was d
ID: 493612 • Letter: G
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.
Explanation / Answer
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
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.