Question

A solution is prepared by mixing 100 mL of 0.120 M HCl and 120 mL of 0.100 M CH3COONa. The pH of the resulting solution is O A. 3.00 O B. 0.92 OC. 2.87 O D. 4.00 O E. 3.42 Reset Selection

Answer 1

mmoles of HCl = 100 x 0.120 = 12

mmoles of CH3COONa = 120 x 0.1 = 12

CH3COONa + HCl --------------> CH3COOH + NaCl

12                   12                                  0               0

0                        0                              12             0

here only CH3COOH is left .

CH3COOH concentration = 12 / 120 + 100 = 0.0545 M

pH = 1/2 [pKa -log C]

pH = 1/2 [4.74 -log 0.0545]

pH = 3.00

answer : A) 3.00

Answer 2

Consider reaction, CH₃COONa + HCl $\rightarrow$ CH₃COOH

Let's calculate mmol of CH₃COONa and HCl .

mmol of CH₃COONa = concentration $\times$ volume = 0.100 M $\times$ 120 ml = 12.0 mmol

mmol of HCl = concentration $\times$ volume = 0.120 M $\times$ 100 ml = 12.0 mmol

Let's use ICE table.

mmol	CH3COONa + HCl $\rightarrow$ CH3COOH
I	12.0	12.0
C	-12.0	-12.0	+12.0
E	0.00	0.00	12.0

After reaction all the CH₃COONa is consumed by added HCl. After reaction solution contain acetic acid and pH of solution is governed by dissociation of acetic acid.

After mixing two solutions, volume of final solution = 100 ml + 120 ml = 220 ml

[ CH₃COOH ] = 12.0 mmol / 220 ml = 0.0545 M

Now, consider dissociation of acetic acid in water.

CH₃COOH ( aq) + H₂O (l) $\rightleftharpoons$ CH₃COO ^- (aq) + H₃O ⁺ (aq)

For above reaction, K a = [ CH₃COO ^- ] [ H₃O ⁺ ] / [ CH₃COOH ]

Assume X moles of acetic acid dissociated at equilibrium, then we can write K a = ( X ) ( X ) / 0.0545 - X

Assume X is very small as compared to 0.0545. Therefore, we can write 0.0545 - X = 0.0545.

$\therefore$ X ² = 0.0545 K a

X ² = 0.0545 ( 1.75 $\times$ 10 ^-05 )

X ² = 9.538 $\times$ 10 ^-08

X = 9.766 $\times$ 10 ^-04 M

We have, X = [ CH₃COO ^- ]= [ H₃O ⁺ ] = 9.766 $\times$ 10 ^-04 M

We have relation, pH = - log [ H₃O ⁺ ]

$\therefore$pH = -log ( 9.766 $\times$ 10 ^-04 )

pH = 3.00

ANSWER : A. 3.00