Find the pH of a 0.002 N acetic acid solution,it is 1.3% ionised at a given dilution
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Hlw mate ,
This is how we go through the calculation of pH.
Concentration 0.002 N
Degree of ionization 2.3%
Then we can go a head by calculating the equilibrium concentration of acetic acid.
Equilibrium concentration = C∝
Where C is the concentration and ∝ is the degree of ionization.
= 0.002 x 2.3/100
= 4.6 x 10^-5 M
Then pH= -log concentration
pH= -log 4.6 x 10^-5
= 4.3.
Hope it helpful
This is how we go through the calculation of pH.
Concentration 0.002 N
Degree of ionization 2.3%
Then we can go a head by calculating the equilibrium concentration of acetic acid.
Equilibrium concentration = C∝
Where C is the concentration and ∝ is the degree of ionization.
= 0.002 x 2.3/100
= 4.6 x 10^-5 M
Then pH= -log concentration
pH= -log 4.6 x 10^-5
= 4.3.
Hope it helpful
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