1) You need to produce a buffer solution that has a pH of 5.33. You already have
ID: 902969 • Letter: 1
Question
1) You need to produce a buffer solution that has a pH of 5.33. You already have a solution that contains 10. mmol (millimoles) of acetic acid. How many millimoles of acetate (the conjugate base of acetic acid) will you need to add to this solution? The pKa of acetic acid is 4.74.
2) Part A: What is the pH of a buffer prepared by adding 0.506 mol of the weak acid HA to0.406 mol of NaA in 2.00 L of solution? The dissociation constant Ka of HA is 5.66×107.
Part B: What is the pH after 0.150 mol of HCl is added to the buffer from Part A? Assume no volume change on the addition of the acid.
Part C: What is the pH after 0.195 mol of NaOH is added to the buffer from Part A? Assume no volume change on the addition of the base.
3) 80.0 mL volume of 0.25 M HBr is titrated with 0.50 M KOH. Calculate the pH after addition of 40.0 mL of KOH.
Explanation / Answer
1) pH = pKa + log [acetic acid] /[acetate]
5.33 = 4.74 + log (10 / [acetate])
10 / [acetate] = 3.89
acetate millimoles = 2.57
2)
pKa = -log Ka = 6.25
pH = pKa + log [NaA] / [HA]
pH = 6.25 + log (0.406 / 0.506)
pH = 7.05
Part B: What is the pH after 0.150 mol of HCl is added to the buffer from Part A
on addition of acid to acidic buffer acid moles increases and salt moles decreases
HCl = C = 0.150 mol
pH = pKa + log [salt -C] / [acid + C]
pH = 6.25 + log (0.406 -0.150 / 0.506 + 0.150)
pH = 5.84
Part C: What is the pH after 0.195 mol of NaOH is added to the buffer from Part A
on addition of base salt moles increases and acid moles decreases
pH = 6.25 + log (0.406 + C / 0.506 -C)
pH = 6.25 + log (0.406 + 0.195 / 0.506 -0.195)
pH = 6.54
3) 80.0 mL volume of 0.25 M HBr is titrated with 0.50 M KOH. Calculate the pH after addition of 40.0 mL of KOH.
HBr millimoles = 80 x 0.25 = 20
KOH millimoles = 0.50 x 50 = 25
base dominates here
base remaining millimoles = 25 -20 = 5
base concnetration = 5 / total volume
= 5 / (80+40)
= 0.042
[OH-] = 0.042 M
pOH = -log [OH-] = 1.38
pH +pOH = 14
pH =12.62
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