Question

A 0.0560-g quantity of acetic acid is dissolved in enough water to make 50.0 mL of solution. Calculate the concentrations of H+, CH3COO−, and CH3COOH at equilibrium. (Ka for acetic acid = 1.8 × 10−5.)

Answer 1

Given : Mass of acetic acid = 0.0560 g

Volume of solution = 50.0 mL = 0.050 L

[CH3COOH ] = mol CH3COOH / volume in L

Mol CH3COOH = mass in g / molar mass

= 0.0560 g / 60.05 g per mol

= 0.000932 mol

[CH3COOH]= 0.000932 mol / 0.050 L

= 0.019 M

ICE

            CH3COOH (aq) + H2O (l)     --- > CH3COO- (aq)     + H3O+ (aq)

I          0.019                                                   0                                  0

C        -x                                                         +x                               +x

E          (0.019 –x)                                           x                                  x

Ka = [CH3COO-][H3O+]/ [CH3COOH]

1.8 E -5 = x² / ( 0.019-x)

Since the value of ka is small we use 5% approximation

(0.019 –x ) = 0.019

1.8 E -5 = x² / 0.019

We find value of x

X² = 1.8 E-5 x 0.019

x = 0.000585

[CH3COO-] = 0.000585 M

[H3O+] = x = 0.000585 M

[CH3COOH]= 0.019 – x = 0.019 – 0.000585 = 0.0184 M