Question

What is the pH when 6.1 g of sodium acetate, NaC2H3O2, is dissolved in 250.0 mL of water? (The Ka of acetic acid, HC2H3O2, is 1.8×10−5.)

Express your answer using two decimal places.

Answer 1

Molar mass of CH3COONa,
MM = 2*MM(C) + 3*MM(H) + 2*MM(O) + 1*MM(Na)
= 2*12.01 + 3*1.008 + 2*16.0 + 1*22.99
= 82.034 g/mol


mass(CH3COONa)= 6.1 g

number of mol of CH3COONa,
n = mass of CH3COONa/molar mass of CH3COONa
=(6.1 g)/(82.034 g/mol)
= 7.436*10^-2 mol
volume , V = 250.0 mL
= 0.25 L


Molarity,
M = number of mol / volume in L
= 7.436*10^-2/0.25
= 0.2974 M

use:
Kb = (1.0*10^-14)/Ka
Kb = (1.0*10^-14)/1.8*10^-5
Kb = 5.556*10^-10
CH3COO- dissociates as

CH3COO-        + H2O   ----->     CH3COOH +   OH-
0.2974                        0         0
0.2974-x                      x         x


Kb = [CH3COOH][OH-]/[CH3COO-]
Kb = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes

Kb = x*x/(c)
so, x = sqrt (Kb*c)
x = sqrt ((5.556*10^-10)*0.2974) = 1.285*10^-5

since c is much greater than x, our assumption is correct
so, x = 1.285*10^-5 M



use:
pOH = -log [OH-]
= -log (1.285*10^-5)
= 4.89


use:
PH = 14 - pOH
= 14 - 4.89
= 9.11

Answer: 9.11