1. Calculate the pH of each of the following strong acid solutions: a) 0.0167 M

Question

1. Calculate the pH of each of the following strong acid solutions:

a) 0.0167 M HNO3, b) 0.225 M HClO3 in 2.00 L of solution, c) 15.00 mL of 1.00 M HCl diluted to 0.500 L, d) a mixture formed by adding to 50.0 mL of 0.020 M HCl to 125 mL of 0.010 M HI.

2. A 0.100 M solution of bromoacetic acid (BrCH2COOH) is 13.2% ionized. Calculate [H^+], [BrCH2COO-], [BrCh2COOH] and Ka for bromoacetic acid.

3. Codeine (C18H21NO3) is a weak organic base. A 5.0 x10^-3 M solution of codeine has a pH of 9.95. Calculate thebalue of Kb for this substance. What is the pKb for this base?

4. Explain the following observation: a) HCl is a stronger acid then H2S, b) H3PO4 is a stronger acid then H3AsO4, c) HBrO3 is a stronger acid then HBrO2, d) H2C2O4 is a stronger acid then HC2O4-, e) benzoic acid (C6H5COOH) is a stronger acid than phenol (C6H5OH).

Explanation / Answer

1, a) pH = - log[H+] = - log (0.0167) = 1.777

b) Concentration of solution =0.225M

0.225 moles are present in 1L of solution

Therefore in 2L of solution (0.225 * 2) = 0.45 moles are present

pH = -log[H+] = -log (0.45) = 0.35

c) Concentration of diluted solution = (15 * 1) / (0.5 * 1000) = 0.03M

pH = -log(0.03) = 1.52

d) 0.02M HCl contains 0.02 moles of H+ in 1000 ml of solution

therefore 50ml of the solution contains 0.001 moles of H+

Similarly 125 ml of 0.01M HI solution contains (0.01/1000)*125 = 0.00125 moles of H+

Therefore 175ml of the mixture contains (0.001 + 0.00125) = 0.00225 moles of H+

pH = -log(0.00225) = 2.65

2. BrCH2COOH ----------------- H+ + BrCH2COO-

13.2% of 0.1M = 0.0132 M

Since the solution is only 13.2% ionized, therefore [H+] = [BrCH2COO-] = 0.0132M

[BrCH2COOH] = 0.1 - 0.0132 = 0.0868M

Ka = ([H+] * [BrCH2COO-])/ ([BrCH2COOH]) =(0.0132 * 0.0132) / 0.0868 = 0.002

3. C18H21NO3 + H2O --------------------- C18H22NO3+ + OH-

Kb = ([C18H22NO3+]*[OH-]) / [ C18H21NO3]

pH + pOH = pKw = 14

pOH = pKw - pH =14 - 9.95 = 4.05

pOH = - log[OH-] =4.05

Therefore [OH-] = 10(-4.05) = 8.9125 x 10-5 M

[ C18H22NO3+] = [OH-] = 8.9125 x 10-5 M (In order to maintain neutrality)

[C18H21NO3] = (5 x 10-3) - (8.9125 x 10-5) = 0.00491M

Kb = {(8.9125 x 10-5 ) * ( 8.9125 x 10-5) } / (0.00491) = 1.62 x 10-6

pKb = -log Kb = - log (1.62 x 10-6) = 5.79

4. a) The conjugate base of HCl is Cl- while that of H2S is HS- . In case of HCl the negative charge is present on the Cl atom while in the second case the negative charge is present on the S atom. Cl is more electronegative as compared to S. Thus Cl- is more stable as compared to HS- . More is the stability of the conjugate base stronger is the acid. Thus HCl is a stronger acid than H2S.

b) Electronegativity decreases down a group. As is less electronegative as compared to P. Thus the conjugate base of H3PO4, that is H2PO4- is more stable than than the conjugate base of H3AsO4. More is the stability of the conjugate base, stronger is the acid. Thus H3PO4 is stronger than H3AsO4 in terms of acidic strength.

c) The conjugate base of HBrO3 is BrO3- while that of HBrO2 is BrO2-. In the former, the negative charge is delocalised over three oxygen atoms while in the latter the negative charge is delocalised over only two oxygen atoms. More is the delocalisation of the negative charge, more is the stability. Thus BrO3- is more stablke than BrO2-. More stable is the conjugate base stronger is the acid. Thus HBrO3 is stronger than HBrO2.

d) The conjugate base of H2C2O4 is HC2O4- which is very stable. Thus H2C2O4 tends to lose it proton easily to form this stable conjugate base. On the other hand HC2O4- would lose a proton to form the conjugate base C2O4- which is less stable.

e) The conjugate base of benzoic acid is the benzoate ion where the negative charge can be esily delocalised over the two oxygen atoms. The conjugate base of phenol is the phenoxide ion where the negative charge is localised over only one oxygen atom. Thus benzoate ion is more stable than phenoxide ion. Thus benzoic acid is more acidic than phenol.

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