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Grade: 12th pass
        
proove that ph of an aqeous solution of a weak base is ph equals pkw-1/2(pkb-logc)
one year ago

Answers : (1)

Arun
22818 Points
							
 

The extent to which hydrolysis proceeds is expressed as the degree of hydrolysisand is defined as the fraction of one mole of the salt that is hydrolysed when the equilibrium has been attained. It is generally expressed as h or x.

h = (Amount of salt hydrolysed)/(Total salt taken)

Considering again eq. (i),

Kh = x2C/(1-x)   or     Kh = h2C/(1-h)

When h is very small (1-h) → 1,

H2 = Kh × 1/c

 or   h = √(Kh/C)  = √(Kw/Kb \timesC)

[H+] = h × C = √(C \times Kh)/Kb

log [H+] =  1/2 log Kw + 1  1/2log C - 1/2log Kb

pH = \frac{1}{2}pKw - \frac{1}{2} log C - \frac{1}{2} pKb = 7 - \frac{1}{2} pKb - \frac{1}{2}log C

 
one year ago
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