Question

In: Chemistry

3.920 g of sodium acetate ( CH3COONa MW= 82.03 g/mol), is dissolved in 235.0ml of H2O...

3.920 g of sodium acetate ( CH3COONa MW= 82.03 g/mol), is dissolved in 235.0ml of H2O at 50 degree celsius (kw= K.48E-14). Calculate the pH of this soluion ( Ka= 1.63 E-5) Determine the % ionization

Expert Solution

When CH3COONa is dissolved in water following reaction occur,

CH3COO^- + H2O ---> CH3COOH + OH^-

Kb = Kw/Ka

= 5.48*10^-14/1.63*10^-5

Kb = 3.36*10^-9

[CH3COONa] = moles of CH3COONa / Volume in L

moles = mass/ molar mass

So, moles of CH3COONa = 3.920g/82.03g/mol = 0.0478mol

So, [CH3COONa] = 0.0478mol/0.235L = 0.2034M

Let [ CH3COOH] = [OH^-] = x

So, Kb = [CH3COOH][OH^-] / [CH3COO^-]

3.36*10^-9= x*x/0.2034M

x² = 6.834*10^-8M

x = (6.834*10^-8)^1/2M

x = 2.614*10^-4M

[OH^-] = 2.614*10^-4M

pOH = -log[OH^-]

= -log(2.614*10^-4)

pOH = -log2.614+4log10

= 3.58

pH = 14-pOH = 14-3.58 = 10.4

pH = 10.4

Percent ionization =( [OH^-] / [CH3COONa] )*100

=( 2.614*10^-4M / 0.2034M)*100

= 0.128%

queen_honey_blossom answered 2 years ago

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