Acid-Base Equilibrium Dry-Lab Exercise Solution Preparation: Calculate the [H_3O

Question

Acid-Base Equilibrium Dry-Lab Exercise Solution Preparation: Calculate the [H_3O^+], [OH] and pH of each of the following solutions: 1.02gof H_2SO_4 diluted to 250.mL of solution (assume a diprotic acid) 2.00 mL of 0.500 M perchloric acid, HCIO4. diluted to 50.0 mL with water (Assume that the volumes add!) 4.50 mL of concentrated sodium hydroxide solution (8.5M) diluted to 12.5L with water 350 mL of an aqueous solution that is 0.20 M in NaCI and 0.15 M in MgCI2. For this question, also calculate [Cl']

Explanation / Answer

(i) concentration of H2SO4 = 1.02 g/98.08 g/mol x 0.250 L = 0.041 M

H2SO4 <==> H+ + HSO4-

First dissociation is 100%, so we have 0.041 M H+ and 0.041 M HSO4-

HSO4- <==> H+ + SO4^2-

let x amount of HSO4- has dissociated, applying this to Ka equation,

1.2 x 10^-2 = x^2/0.041

x = [H+] = 0.022 M

total [H+] = 0.041 + 0.022 = 0.063 M

pH = -log[H+] = -log(0.063) = 1.20

[OH-] = 1.58 x 10^-13 M

(ii) concentration of diluted HClO4 solution = 0.5 M x 2 ml/50 ml = 0.02 M

[H+] = 0.02 M

pH = -log[H+] = 1.70

[OH-] = 5.01 x 10^-13 M

(iii) concentration of dilute NaOH solution = 8.5 M x 4.50 ml/12.5 x 1000 ml = 3.06 x 10^-3 M

[OH-] = 3.06 x 10^-3 M

pOH = -log[OH-] = 2.51

pH = 14 - pOH = 11.49

[H+] = 3.23 x 10^-12 M

(iv) 0.2 M NaCl will have 0.2 M Cl-

NaCl and MgCl2 are strong electrolytes, completely dissociate

[H+] = [OH-] = 1 x 10^-7 M

pH = 7

0.15 M MgCl2 has 0.30 M Cl-

So total [Cl-] = 0.50 M

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