What minimum volume of a buffer solution which is 0.1 M in both lactic acid
(Kº= 1.4 x 104) and sodium lactate must be diluted to exactly 100 cm so that the
resulting buffer solution will change its pH by not more than 0.5 pH unit upon the
addition of 10.0 cm of 0.01 M HCI?
Answers
Answer:
TUM, TA, TUM, TUM How many moles (and grams) of HCl (or stomach acid) could be neutralized by 0.75 g of the antacid CaCO3?
CaCO3 + H2O Ca2+(aq) + 3H2O(l) + CO2(g)
CaCO3 = 0.75 g / 100.1 g/mol = 7.49 x 10-3 mol
HCl = 2 x 7.49 x 10-3 mol = 1.5 x 10-2 mol
1.5 x 10-2 mol x 36.46g/mol = 0.547 g HCl
2. Vinegar is a 5 % (by weight) aqueous solution of H3CCOOH (acetic acid).
i) what is the pH of the solution that you put on your french fries (claiming you don't use vinegar on your fries is not an acceptable answer to this question). ii) What will happen if you try and make a 50% solution of H3C(CH2)4COOH in water? iii) Why? This compound, caproic acid, is a constituent of goat sweat (giving it, in part, the lovely fragrance). You may not want to add this to your french fries.
i) A 5% solution of CH3COOH = 50 g/L. The molecular weight of acetic acid is 60. A 5% solution is therefore a 0.83 M solution.
Ka = 1.8 x 10-5 = [CH3COO-][H3O+]/[CH3COOH] = x2 / 0.83-x Assume x~0 x2 = 1.49 x 10-5
x = [H3O+] = 3.87 x 10-3 (x is < 5% 0.83) pH = 2.41
ii) Caproic acid is much less soluble in water than acetic acid and it is not possible to make a 50% solution.
iii) The London forces between the organic part (CH2)5CH3 are sufficiently strong that the energetic cost of cost of mixing with water are too high. Similarly, the loss of the hydrogen bonding of water with itself when it mixes with the caproic acid is too high. The system remains in two phases.
3. Suppose you want (or I force you) to prepare a buffer solution to maintain the pH at 4.30. Which of the three acids given below would be appropriate? What ratio of the acid/conjugate base should be used to fix the pH at 4.3?
Acid
HSO4-
CH3COOH
HCN Conjugate
HSO4-
CHCOO-
CN-SO42- Base
1.2 x 10-2
1.8 x 10-5
4.0 x 10-10
The desired pH is 4.3, therefore the [H3O+] is 10-4.3 = 5 x 10-5. Only acetic acid has a Ka close to this so it is the only possible choice.
Henderson Hasselblad: pH = pKa + log ([A-]/[HA]) 4.3 = 4.74 + [A-]/[HA]
log ([A-]/[HA])= -0.44
[A-]/[HA]= 0.36, i.e., 1 M HOOCCH3 and 0.36 M NaOOCCH3
4. Derive the relationship below for a given acid/conjugate base pair: (For convenience use HA as your general acid.)
pKa + pKb = pKw
HA + H2O H3O+ + A-
Keq = [H3O+][A-]/[H2O][HA]
Keq [H2O] =Ka = [H3O+][A-]/[HA]
pKa = pH -log([A-]/[HA])
A- + H2O OH- + HA
Keq = [HA][OH-]/[A-] [H2O]
Keq [H2O] =Kb = [HA][OH-]/[A-]
pKb = pOH -log([HA]/[A-])
Kw = [H3O+][OH-]
pKw = pH + pOH
pKa + pKb = pH -log([A-]/[HA]) + pOH -log([HA]/[A-])
pKa + pKb = pH + log([HA]/[A-]) + pOH -log([HA]/[A-]) = pH + pOH = pKw
5. Calculate the pH of a 0.10 M solution of triethylamine hydrochloride in water. Ka (H3CCH2)3NHCl = 7.94 x 10-11 at 25 oC.
x2 /(0.1-x) = Kb assume x << 0.1
x2/0.1 = 7.94 x 10-12
x = 2.81 x 10-6 x is << 0.1
pH = 5.55
6. A buffer solution is prepared by mixing equal volumes of 0.100 M lactic acid and 0.200 M sodium lactate (Ka = 1.4 x 10-4). i) Calculate the pH of the resulting solution. ii) What will the pH be after adding 10 mL 0.2 M HCl to a 100 mL solution of the lactate buffer?
Dilute each compound by mixing (1:1). Thus, concentration lactic acid = 0.05 M, of sodium lactate = 0.1 M
i) Henderson Hasselbalch pH = pKa + log [A-]/[HA]
pH = -log (1.4 x 10-4 ) + log [0.1]/[0.05] = 3.85 + 0.30 = 4.15
ii) HCl = 10 mL x 0.0002 mol/mL = 0.002 mol This will decrease the sodium lactate concentration
number moles sodium lactate = 0.100M x 0.100 L = 0.01 mol,
moles lactic acid = 0.05 M x 0.100 L = 0.005 mol
After addition 0.002 mol HCl, number moles sodium lactate = 0.100M x 0.100 L = 0.01 mol-0.002 mol = 0.008 mol,
moles lactic acid = 0.05 M x 0.100 L = 0.005 mol + 0.002 mol = 0.007 mol
New volume = -0.11 L
pH = pKa + log [0.008 mol/0.11 L)]/[0.007 mol/ 0.11L]
pH = 3.85 + 0.058 = 3.91
7. How many grams of NH4Br(s) must be added to 1.50 L of 0.30 M NH3 solution to give a pH of 8.74? Assume there are no volume changes. Kb NH3 = 1.8 x 10-5.
Henderson Hasselbalch pOH = pKb + log [HA]/[A-], and pOH = 14-pH, pOH = 14 - 8.74 = 5.26
= 4.74 + log [HA]/[A-]
0.52 = log [HA]/[A-]
3.31 = [HA]/[A-]
3.31 = [HA mol/ L]/[0.3 mol/L]
1 = [HA mol/L] Therefore need to add 1 mol of NH4Br /L - have 1.5 L, therefore must add 1.5 mol. MW NH4Br = 98
98 x 1.5 = 147 g