Question

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CH,CH,COOH(aq) + H2O(1) = CH3CH2COO(aq) + H30*(aq) Propanoic acid, CH,CH,COOH, is a carboxylic acid that reacts with water according to the equation above. At 25°C the pH of a 50.0 mL sample of 0.20 M CH,CH,COOH is 2.79. a. Identify a Brønsted-Lowry conjugate acid-base pair in the reaction. Clearly label which is the acid and which is the base. b. Determine the value of K, for propanoic acid at 25°C.

Answer 1

a) CH3CH2COOH + H2O -----------------------> CH3CH2COO- + H3O+

acid base conjugate base conjugate acid

CH3CH2COOH and CH3CH2COO- is one conjugate acid-base pair.

b) propanoic acid is a weak acid

the pH of weak acid is given by

pH = 1/2[pKa -log C]

Given pH = 2.79

C = 0.20M

Then

2.79 = 1/2 [ pKa -log0.20]

solvin g pKa = 4.881

Ka = 10^-4.881 = 1.315x10^-5

c)

i)CH3CH2COOH + NaOH -----------------------> CH3CH2COONa + H2O

50x0.2 50x0.2 0 0 initial mmoles

0 0 50x0.2 -- after rxn

The solution has a salt of weak acid and strong base, so will be basic in solution with pH >7.

FALSE.

ii) HCl is a strong acid dissociates completely but popanoic acid is weak dissociates to small extent,

Thus [H3O+} from HCl is much higher than from propanoic acid of same concentration.

Thus if both have to have same pH ,that is same [H3O+] the concentration of weak acid should be much less , so that the degree of dissociation is equal to 1 .

FAlse.

d)

For titration

mmoles of acid = mmoles of base

V1M1 = V2M2

25.00mL x M1 = 0.173 x 20.52

M1 = 0.142 M