1.A buffer solution contains 0.241 M Na H 2 PO 4 and 0.318 M K 2 HPO 4 . If 0.03
ID: 304381 • Letter: 1
Question
1.A buffer solution contains 0.241 M NaH2PO4 and 0.318 M K2HPO4.
If 0.0308 moles of hydroiodic acid are added to 150. mL of this buffer, what is the pH of the resulting solution ?
(Assume that the volume does not change upon adding hydroiodic acid)
2. A buffer solution contains 0.322 M nitrous acid and 0.492 M potassium nitrite.
If 0.0359 moles of hydrochloric acid are added to 150 mL of this buffer, what is the pH of the resulting solution ?
(Assume that the volume does not change upon adding hydrochloric acid)
3. A buffer solution contains 0.308 M KHSO3 and 0.293 M K2SO3.
If 0.0380 moles of sodium hydroxide are added to 225 mL of this buffer, what is the pH of the resulting solution?
(Assume that the volume does not change upon adding sodium hydroxide)
Explanation / Answer
Henderson–Hasselbalch equation for weak acid/base is given by
pH= pKa +log [A-]/[HA], [HA] is the concentration of undissociated weak acid, [A?] is the concentration of this acid's conjugate base.
(1)
Concentrations,
No. of moles of NaH2PO4 (HA) in 0.241 M, 150 mL = 0.241 M x 150 mL /1000 mL = 0.03615 mol
No. of moles of K2HPO4 (A-) in 0.318 M, 150 mL = 0.318 M x 150 mL /1000 mL = 0.04770 mol
No. of moles of HI added = 0.0308 mol.
On adding 0.0308 mol of HI, 0.0308 mol of HPO4^2- (A-) is converted to 0.0308 mol of H2PO4^- (HA).
So,
the final concentration of (HA) = 0.03615 mol + 0.0308 mol = 0.06695 mol.
the final concentration of (A-) = 0.04770 mol - 0.0308 mol = 0.01690 mol.
the final volume of the mixture =150 mL +0 mL = 150 mL
(Since volume is not changing you can also directly use no. of mol, instead of converting to molarity, but I am converting to molarity)
Final molarity of (HA) in buffer = 0.06695 x 1000 mL/150 mL = 0.4463 M
Final molarity of (A-) in buffer = 0.01690 x 1000 mL/150 mL = 0.1127 M
Using Henderson–Hasselbalch equation,
Ka of phosphoric acid (2) = 6.23 x 10^-8, pKa = -log Ka = -log ( 6.23 x 10^-8) = 7.21
pH= pKa +log [A-]/[HA] = 7.21 + log[0.1127 M/0.4463 M] = 7.21 - 0.60 = 6.61
pH of buffer after adding hydroiodic acid = 6.61
------------------------------------------------------------------------------------
(2)
Concentrations,
No. of moles of HNO2 (HA) in 0.322 M, 150 mL = 0.322 M x 150 mL /1000 mL = 0.0483 mol
No. of moles of KNO2 (A-) in 0.492 M, 150 mL = 0.492 M x 150 mL /1000 mL = 0.0738 mol
No. of moles of HI added = 0.0359 mol.
As above,
the final concentration of (HA) = 0.0483 mol + 0.0359 mol = 0.0842 mol.
the final concentration of (A-) = 0.0738 mol - 0.0359 mol = 0.0379 mol.
the final volume of the mixture =150 mL +0 mL = 150 mL
Final molarity of (HA) in buffer = 0.0842 x 1000 mL/150 mL = 0.5613 M
Final molarity of (A-) in buffer = 0.0379 x 1000 mL/150 mL = 0.2527 M
Using Henderson–Hasselbalch equation,
Ka of nitrous acid = 4.0 x 10^-4, pKa = -log Ka = -log ( 4.0 x 10^-4) = 3.39
pH= pKa +log [A-]/[HA] = 3.39 + log[0.2527 M/0.5613 M] = 3.39 - 0.35 = 3.04
pH of buffer after adding hydrochloric acid = 3.04
------------------------------------------------------------------------------------
(3)
Concentrations,
No. of moles of KHSO3 (HA) in 0.308 M, 225 mL = 0.308 M x 225 mL /1000 mL = 0.0693 mol
No. of moles of K2SO3 (A-) in 0.293 M, 225 mL = 0.293 M x 225 mL /1000 mL = 0.0659 mol
No. of moles of NaOH added = 0.0380 mol.
On adding 0.0380 mol of NaOH, 0.0380 mol of KHSO3 (HA) is converted to 0.0380 mol of K2SO3 (A-).
the final concentration of (HA) = 0.0693 mol - 0.0380 mol = 0.0313 mol.
the final concentration of (A-) = 0.0659 mol + 0.0380 mol = 0.1039 mol.
the final volume of the mixture = 225 mL + 0 mL = 225 mL
Final molarity of (HA) in buffer = 0.0313 x 1000 mL/225 mL = 0.1391 M
Final molarity of (A-) in buffer = 0.1039 x 1000 mL/225 mL = 0.4618 M
Using Henderson–Hasselbalch equation,
Ka of Sulphurous acid (2) = 6.6 x 10^-8, pKa = -log Ka = -log (6.6 x 10^-8) = 7.18
pH= pKa +log [A-]/[HA] = 7.18 + log[0.4618 M / 0.1391 M] = 7.18 + 0.52 = 7.70
pH of buffer after adding NaOH = 7.70
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.