Question

Calculate the ph of 0.1 m solution of sodium acetate

Answer 1

CH3COONa is a strong salt : CH3COO- + Na+ 

CH3COO- + H2O <------> CH3COOH + OH- 

Kb = Kw/Ka = 5.6 x 10^-10 = x^2/ 0.1-x 

x = 7.5 x 10^-6 M = [OH-] 

pOH = 5.1 

pH = 14 - 5.1 = 8.9

Answer 2

Concept:

The ph is calculated as

pH equals -log [H+] is the hydrogen ion concentration in moles per litre solution, where log is the base-10 logarithm and [H+] is the hydrogen ion concentration in moles per litre solution. 

Given:

$0.1m$ solution of sodium acetate.

To Find:

ph of $0.1m$ solution of sodium acetate.

Solution:

The concentration of hydrogen ions is calculated to determine the pH.

The degree of hydroxylsis for sodium acetate is $7.48 \times 10^{-5}$

The hydroxyl ion concentration will be:

$\left[O H^{-}\right]=c h=0.1 \times 7.48 \times 10^{-5}=7.48 \times 10^{-6} M$

Concentraton of Hydrogen ions-

$\left[H^{+}\right]=\frac{K_{w}}{\left[O H^{-}\right]}$

$K_{w} = 10^{-14}$, ionic product of water

$\left[H^{+}\right]=\frac{10^{-14}}{7.48 \times 10^{-6}}=1.33 \times 10^{-9} M$

Calcutating the ph:

$p H=-\log \left[H^{+}\right]=-\log \left(1.33 \times 10^{-9}\right)=8.88$

The ph of $0.1m$ solution of sodium acetate is 8.88.

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