Any important information given in the problem; If you use information, such as

Question

Any important information given in the problem; If you use information, such as a K_a from one of the tables in your book or the internet, you must reference where you found that information. Any chemical equation necessary for working the problem. Mathematical substitutions into the formulae and subsequent arithmetic and mathematical simplification; The final answer is clearly marked. With the aid of Table C, predict the direction (forward or reverse) favored in each of the following acid-base reactions. (a) CH_3COOH + CO_3^2- HCO_3^- + CH_3 CO (b) HNO_2 + ClO_4^- HClO_4 + NO_2^- (c) H_2CO_3 + CO_3^2- HCO_3_3^- + HCO_3^- What is [H_3 O^+] in a solution obtained by dissolving 205 mL HCl(g), measured at 23 degree C and 751 mmHg, in + .25 L of aqueous solution? Caproic acid, HC_5H_11O_2, found in small amounts in coconut and palm oils, is used in making artificial flavors. A saturated aqueous solution of the acid contains 11 g/L and has pH = 2.94. Calculate K_2 for the acid. HC_ H_ O_2 + H_2O H_ O^- + C_6H_11O_2^- K_a = ? What are [H_3O^-], [OH^-], pH, and pOH of 0.386 M CH_3NH_2?

Explanation / Answer

1. direction of the reactions based on pKa and pKb

(a) CH3COOH is a stronger acid than HCO3- and hence the direction of reaction would be towards right hand side, towards product end.

(b) HClO4 is a stronger acid than HNO2, thus the direction of reaction would be towards left hand side, towards reactant end.

(c) CO3^2- is a stronger base than HCO3- and hence the direction of reaction would be towards right hand side, toards product end.

2. moles of acid = PV/RT

                          = (751/760)(0.205)/(0.08205)(273 + 23)

                          = 0.00834 mol

So,

[H3O+] concentration = moles/L of solution

                                   = 0.00834/4.25

                                   = 0.00196 M

3. pH = -log[H3O+] = 2.94

[H3O+] = 1.15 x 10^-3 M

[H3O+] = [C5H11COO-]

initial molar concentration of caproic acid solution = 11/116.16 = 0.095 mol/L

Ka = [C5H11COO-][H3O+]/[C5H11COOH]

      = (1.15 x 10^-3)^2/(0.095 - 1.15 x 10^-3)

      = 1.41 x 10^-5

4. 0.386 M CH3NH2

CH3NH2 + H2O <==> CH3NH3+ + OH-

let x amount has reacted

Kb = [CH3NH3+][OH-]/[CH3NH2]

4.4 x 10^-4 = x^2/(0.386 - x)

x^2 + 4.4 x 10^-4x - 1.7 x 10^-4 = 0

x = [OH-] = 0.013 M

pOH = -log[OH-] = 1.89

[H3O+] = 1 x 10^-14/0.013 = 7.80 x 10^-13 M

pH = -log[H3O+] = 12.11

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