a) The dissociation of propanic acid (CH_3 CH_2 CO_2 H) in water b) The dissocia

Question

a) The dissociation of propanic acid (CH_3 CH_2 CO_2 H) in water b) The dissociation of perchloric (HClO_) acid in water Fill the missing information in the following table. Calculate the percent dissociation and pH of the and in each of the following Solutions: Hint, acetic acid is HC_2 H_3 O_2 a) 0.50 M acetic acid b) 0.050 M acetic acid c) 0.0050 M acetic acid d) Even though the percent dissociation increase from solution a to c the [H^+] decrease. Explain If assumption is no good, stop and state why it's as good

Explanation / Answer

1. (a) Dissociation equation for propanoic acid in water:

     CH3CH2COOH(aq) + H2O(l) <=> CH3CH2COO- (aq) + H3O+ (aq)

Or CH3CH2COOH(aq) <=> CH3CH2COO- (aq) + H+ (aq)

(b) Dissociation equation for perchloric acid in water

   HClO4(aq) + H2O(l) <=> ClO4- (aq) + H3O+ (aq)

2. Solution a :

pH = 6.88, pOH = 14-pH = 14 – 6.88 => 7.12

[H+] = 10-pH = 10-6.88 = > 1.32 x 10-7M

[OH-] = 10-pOH = 10-7.12 => 7.58 x 10-8M

pH < 7 thus Acidic

Solution b:

[OH-] = 8.4 x 10-14M, then [H+] = 1 x 10-14/8.4 x10-14 => 0.119M

pH = -log[H+] = -log(0.119) => 0.924

pOH = 14-pH = 14 – 0.924=> 13.076

pH < 7 thus Acidic

Solution c :

pH = 14-pOH = 14 – 3.11=> 10.89

[H+] = 10-pH = 10-10.89 = > 1.288 x 10-11M

[OH-] = 1 x 10-14/1.288 x10-11 => 7.76 x 10-4M

pH > 7 thus Basic

Solution D.

[H+] = 1.0 x 10-7M, pH = -log[H+] = -log(1.0 x 10-7) => 7

[OH-] = 1 x 10-14/1.0 x10-7 = 1.0 x 10-7M

pOH = 7

pH = 7 thus its neutral

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