Question

Acid-Base Chemistry: Unknown Acid Analysis (in two experiments) Introduction In this experiment you will titrate a monoprotic weak acid with a strong ak acid with a strong base in the nd using a pH meter (Exp 2). An analysis of concentration and molar mass of the the titration data will allow you to determine the concentration and more unknown acid (Exp 1) and the ionization constant Ka (Exp 2). At the equivalence point of a titration, exactly enough OH ions have beer y enough OH-ions have been added to react y present. Around this point, the pH of the ne rapid increase in pH can be detected mixture increases suddenly and dramatically. The rapid increase in pH can using an indicator (such as phenolphthalein) or with a pH meter. If you measu amount of OH needed to reach the equivalence point, you also know the more valence point, you also know the moles of HA that were present originally. The halfway point of a titration is halfway to the equivalence point. Enough OH tons have been added to react with exactly half of the HA molecules that were originally present, converting them into A. At this point, there are equal amounts of HA and A in the solution. At the halfway point, the Ka expression simplifies to: [H20+][A-] 1 = [H3O+] since [HA] = [A-] [HA] If you measure the pH at the halfway point, you can easily calculate [H3O+] and Ka for the acid, which should most accurately reflect the strength of your unknown acid. Ka = The halfway point lies in the middle of the buffer region of the titration curve. In the buffer region, there are appreciable concentrations of both HA and A-ion the solution. While the concentrations are equal at the halfway point, they are within the same order of magnitude throughout the buffer region. Pre-Lab Information A4.947 g sample of boric acid was dissolved in enough water to prepare a 100.00 mL solution. The pH titration curve below (next page) was obtained by titrating 25.00 mL of this solution with a 1.000 M NaOH solution. The information given on the titration curve may be needed to answer some of the following questions:

A thorough explanation would sure be appreciated!!!! Thank you!!

Answer 1

Reactions involved are:

B(OH)3 + 2H2O $\rightarrow$ [B(OH)4]- + H3O+

H3O+ + OH-  $\rightarrow$ 2H2O  

Thus boric acid acts as monobasic acid. and can be represented as:

HA + H2O $\rightleftharpoons$ H3O+ + A-

Ka = [H3O+][A-]/[HA]

at any point of titration [HA]o = [HA] + [A-] and [H3O+] = [A-]

at half equivalence point: [A-] = 1/2[HA]o thus [HA] = 1/2[HA]o

upon substitution in Ka expression we get: Ka = [H3O+]

Relating pH and Ka by using the relation, pH = -log [H3O+]

Or, [H3O+] = 10-pH

From the graph we know: haf equivalence Volume of 1 M NaOH = 10 mL and pH = 9.137

thus [H3O+] = 10-9.137

[H3O+] = 7.27*10-10 M

and therefore Ka = 7.27*10-10

Initial concentration of boric acid:

At equivalence point [NaOH] = [HA]o

Volume of 1 M (= M1) NaOH used, V1 = 20 mL

Volume of boric acid used, V2 = 25 mL

Uisng the relation M1V1 = M2V2 we get

Initial concentration of boric acid = 1*20/25 = 0.8 M

From the graph: Calculating Ka at one-quater point: [A-] = 1/4[HA]o =  thus [HA] = 3/4[HA]o

upon substitution in Ka expression we get: Ka = [H3O+] /3

pH = 8.5

thus, [H3O+] = 10-pH

Or, [H3O+] = 10-8.5 = 9.48*10-9 M

thus Ka = 9.48*10-9 /3 = 3.16*10-9

From the graph: Calculating Ka at three-quater point: [A-] = 3/4[HA]o =  thus [HA] = 1/4[HA]o

upon substitution in Ka expression we get: Ka = 3[H3O+]

pH = 9.5

thus, [H3O+] = 10-pH

Or, [H3O+] = 10-9.5 = 3.16*10-10 M

thus Ka = 3.16*10-10 *3 = 9.48*10-10

Average Ka = (7.27*10-10 + 3.16*10-9 + 9.48*10-10)/3 = 1.611 *10-9

2. Ka at equivalence point:

The solution at equivalence point is mainly salt of weak acid as NaA and there is no excess of either weak acid or strong base.

concentration of A- at equivalence = Molarity of NaOH* volume of NaOH used/ total volume of mixture = 20*1/(20+25) = 0.44 M

and now equillibrium is due to NaA as:

	NaA +H2O	OH-	HA
Initial	0.44	0	0
change	-x	+x	+x
equillibrium	0.44-x	x	x

Kb = [HA][OH-]/[A-]/ = x2/0.44-x = $\approx$ x2/0.44

using the relation of Ka and Kb we get:

Kb = Kw/Ka = 10-14/7.27*10-10 = 1.37*10-5

thus x2/0.44 = 1.37*10-5

x = (1.37*10-5 *0.44)1/2 = 2.45*10-3 M

thus [OH-] = x = 2.45*10-3 M

pOH = -log [OH-] = -log (2.45*10-3) =2.61

pH = 14-pOH = 11.39

and equillibrium constant is Kb = 1.37*10-5 and Ka = 7.27*10-10

c) Boric acid is weaker acid than acetic acid as the later has higher Ka about 1.75*10-5. Also acetic acid gives H+

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