Question

do not know where to start on part c and d

11. Propanoic acid, HC3H,O2, ionizes in water according to the equation below K-1.34 X 10 + H3O HCH,O HO CsHsO2( a Write the equilibrium constant expression for the reaction. CHCgHO2 b. Calculate the pH of a 0.265M solution of propanoic acid. 2 Cafleo HO HC,H02H,0 I 0,265M 2 1.34 x (0 L0,265 OM OM dre 2.05x10 tx X 4 0,265-x Aog [.8x 10-M 34x 10-5 X 0,205 veglig 0.00188M-X=[H,0 or 1.88 x 10-M PH=- pH=2.726 c. The methanoate ion, HCO2 (ag) reacts with water to form methanoic acid and hydroxide ion, as shown in the following equation. HCO2(a)+H2Om HCO2Ha OH(a Given the [OH] is 4.18 X 10 M in a 0.309 M solution of sodium methanoate, calculate each of the following. i. The value of K for the methanoate ion, HCO2 (ag ii. The value of K, for methanoic acid, HCO2H. i. Which acid is stronger, propanoic acid or methanoic acid? Justify your answer.

Answer 1

Given reaction is HCO ₂^- (aq) + H₂O (l) $\rightleftharpoons$ HCO ₂H (aq) + OH ^- (aq)

For above reaction, equilibrium constant is K b = [HCO ₂H] [OH ^- ] / [HCO ₂^-]

We have given information that, [OH ^- ] = [HCO ₂H] = 4.18 $\times$ 10 ^-06 M & [HCO ₂^-] = 0.309 M

$\therefore$ K b = (4.18 $\times$ 10 ^-06 ) (4.18 $\times$ 10 ^-06 ) / 0.309

K b = 5.65 $\times$ 10 ^-11

ANSWER : K b of HCO ₂^- = 5.65 $\times$ 10 ^-11

II)  HCO ₂H and HCO ₂^- are conjugate acid base pairs.

We know that, for conjugate acid base pair K a $\times$ K b = K w

$\therefore$ K a of  HCO ₂H = K w / K b of HCO ₂^-

$\therefore$ K a of  HCO ₂H = ( 1 $\times$ 10 ^-14 ) / 5.65 $\times$ 10 ^-11 = 1.77 $\times$ 10 ^-04

ANSWER : K a of  HCO ₂H =  1.77 $\times$ 10 ^-04

d) One compare acidity of acids on the basis of K a value. Larger the K a value more will be acidity and vice versa.

For methanoic acid K a = 1.77 $\times$ 10 ^-04 and for propanoic acid K a = 1.34 $\times$ 10 ^-05 .

K a of methanoic acid is greater than K a of propanoic acid, hence methanoic acid is stronger than propanoic acid.

We have formula, Strength of acid 1 / strength of acid 2 = ( K a 1 / K a 2) ^1/2

$\therefore$ Strength of  methanoic acid / strength of propanoic acid = ( 1.77 $\times$ 10 ^-04 / 1.34 $\times$ 10 ^-05 ) ^1/2 = 3.63

i e Strength of  methanoic acid = 3.63 $\times$ strength of propanoic acid

ANSWER : Methanoic acid is 3.63 times stronger than propanoic acid.