1. Which of the following would make a buffer solution? Explain your choice brie
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Question
1. Which of the following would make a buffer solution? Explain your choice briefly.
(a) Solution A: 45 mL of 0.30 M NaC2H3O2 + 25 mL of 0.17 M NaCl
(b) Solution B: 26 mL of 0.14 M HC2H3O2 + 40 mL of 0.17 M NaOH
(c) Solution C: 26 mL of 0.19 M NaC2H3O2 + 20 mL of 0.066 M HCl
(d) Solution D: 35 mL of 0.11 M HC2H3O2 + 20 mL of 0.25 M NaCl
2. Predict the pH region in which each of the following buffers will be effective, assuming equal molar concentrations of the acid and its conjugate base (refer to Tables in text); (a) sodium lactate and lactic acid; (b) sodium benzoate and benzoic acid; (c) potassium hydrogenphosphate and potassium dihydrogenphosphate.
3. (a) What must be the ratio of the concentrations of CO3 2- and HCO3 - ions in a buffer solution having a pH of 11.0?
(b) What mass of K2CO3 must be added to 1.00 L of 0.100 M KHCO3(aq) to prepare a buffer solution with a pH of 11.0?
4. Calculate the pH of 25.0 mL of 0.110 M aqueous lactic acid being titrated with 0.150 M NaOH(aq) [See text for Ka value.]
(a) initially;
(b) after the addition of 5.0 mL of base;
(c) after the addition of a 5.0 mL more of base, Vt =10.0mL;
(d) at the equivalence point;
(e) after the addition of 5.0 mL of base beyond the equivalence point;
(f) after the addition of 10.0 mL of base beyond the equivalence point;
(g) Pick a suitable indicator from your text, or your Laboratory Manual.
5. Captain Kirk, of the Starship Enterprise, has been told by his superiors that only a chemist can be trusted with the combination to the safe containing the dilithium crystals that power the ship. The combination is the pH of solution A, described below, followed by the pH of solution C. (Example: If the pH of solution A is 3.47 and that of solution C is 8.15, then the combination to the safe is 3-47-8-15.) The chemist must determine the combination using only the information below (all solutions are at 25 oC):
- Solution A is 50.0 mL of a 0.100 M solution of the weak monoprotic acid HX.
- Solution B is a 0.0500 M solution of the salt NaX. It has a pH of 10.02.
- Solution C is made by adding 15.0 mL of 0.250 M KOH to solution A.
What is the combination to the safe?
6. Consider the reaction of lactic acid with ammonia, a weak acid-weak base reaction. To 50.0 mL of a solution containing 0.100 M lactic acid, HLac, 0.100 M NH3(aq) is added until all of the acid has been neutralized.
(a) Is the solution acidic, basic, or neutral at the equivalence point?
(b){Optional;wa/wb, Challenging!} What is the pH at the equivalence point?
Explanation / Answer
1)
Buffer: Buffer is the aqueous solution of any weak acid/base with its conjugate base/acid respectively.
An acid or base must be weak and hence there will be an equilibrium between the amount of acid ionized and unionized (Or in base case amount of base protonated or non-protonated)
Buffer pH maintaining phenomenon based on common ion effect and Le Chatelier’s principle. Salt of weak acid is added to the aq. Solution of weak acid. Salt of weak acid being strong electrolyte ionizes to complete extent and solution contain excess of conjugate base. This excess conjugate base suppress the ionization of acid. Moreover added acid i.e. H+ ions are consumed by conjugate base and get converted into the unionized acid form and no change in pH is observed. Hence the buffer mechanism. Similar way we can explain Basic buffer mechanism.
NaCl is a salt of strong acid (HCl) and strong base (NaOH) hence aqueous solution of NaCl will be neutral and hence no excess of conjugate base present, no consumption of added acid or dilution and hence pH of solution do chages.
Hence option-(a) and (d) are rulled out
Possibility of Buffer formation is with (b) or (c)
As explained earlier, for the buffer solution conjugate partner (In acid case conjugate base and in base case conjugate acid) must be in far excess.
Let calculate milimoles of species in each case,
(b) Solution B: 26 mL of 0.14 M H3CCOOH + 40 mL of 0.17 M NaOH
Milimoles of H3CCOOH = Molarity of H3CCOOH x Volume of H3CCOOH = 0.14 x 26 = 3.64 milimoles
Milimoles of NaOH = Molarity of NaOH x Volume of NaOH = 0.17 x 40 = 6.8 milimoles
Milimoles of Base NaOH > Milimoles of H3CCOOH
NaOH is strong base and will completely neutralize the acid and remaining OH ions will make solution basic.
(c)
Solution C: 26 mL of 0.19 M NaC2H3O2 + 20 mL of 0.066 M HCl
Milimoles of NaOOCCH3 = 0.19 x 26 = 4.94 milimoles
Milimoles of HCl = 0.066 x 20 = 1.32 milimoles
1.32 milimoles of HCl will convert 1.32 milimoles of NaOOCCH3 into HOOCCH3 and still solution contain 4.94 – 1.32 = 3.62 milimoles of NaOOCCH3 i.e. 3.62 milimoles of conjugate base -OOCCH3 and hence we have,
Milimoles of conjugate base -OOCCH3 > Milimoles of HOOCCH3
Hence
Solution C: 26 mL of 0.19 M NaOOCCH3 + 20 mL of 0.066 M HCl will form a buffer,
Let us calculate pH of this buffer,
Accoring to Henderson’s equation,
pH = pKa + log{[conjugate base]/[Acid]}
We have,
pKa = 4.756 for acetic acid,
[conjugate base] = [-OOCCH3] = 3.62 milimoles
[Acid] = [HOOCCH3] = 1.32 milimoles
Let us put all these values in above Henderson’s equation and calculate pH,
pH = 4.756 + log(3.62/1.32)
pH = 4.756 + 0.438
pH = 5.194
Hence the acidic buffer formation.
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2)Generally the pH range of buffer is within the value, 1 unit below pKa or pKb to the 1 unit above pKa or pKb.
a) Sodium lactate/Lactic acid :
pKa of Lactic acid = 3.86
Buffer range for Sodium lactate/Lactic acid buffer is 2.86 – 4.86 pH unit
b) Sodium benzoate/Benzoic acid:
pKa of Benzoic acid = 4.20
Buffer range for Sodium benzoate/Benzoic acid buffer is 3.20 – 5.20 pH unit.
c) KH2PO4/K2HPO4:
pKa of KHPO4 is 6.85
Buffer range for KH2PO4/K2HPO4 buffer is 5.85 – 7.85 pH unit.
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3)
By Henderson’s equation, pH of buffer is given as,
pH = pKa + log{[conjugate base]/[Acid]}
3a) We have pKa of HCO3- = 10.32
pH = 11
We want [CO32-]/[HCO3-] = ?
Let us put pH and pKa values in Hendersons equation above,
11 = 10.32 + log([CO32-]/[HCO3-])
log([CO32-]/[HCO3-]) = 11 – 10.32
log([CO32-]/[HCO3-]) = 0.68
[CO32-]/[HCO3-] = 100.68
[CO32-]/[HCO3-]) = 4.786
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3b) K2CO3 and KHCO3 forms buffer system depicted above as,
K2CO3 ---------> 2K+ + CO32-
KHCO3 ---------> K+ + HCO3-
Hence for this buffer at pH =11 we have,
[CO32-]/[HCO3-]) = 4.786
i.e.
[K2CO3]/[KHCO3]) = 4.786
[K2CO3] = 4.786 x [KHCO3]
MK2CO3 x VK2CO3 = 4.786 x MKHCO3 x VKHCO3
With given things and assuming no volume change,
MK2CO3 x 1.0 L = 4.786 x 0.1 M x 1.0 L
MK2CO3 = 4.786 x 0.1
MK2CO3 = 0.4786 moles of K2CO3
Molar mass of K2CO3 = 138.205 g/mol
I.e. 1 mole of K2CO3 = 138.205 g
So, 0.4786 moles of K2CO3 = 0.4786 x 138.205 = 66.145 g of K2CO3
66.145 g of K2CO3 needed to be added to make buffer of pH = 11.
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