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prove that pH=1/2[pKw+pKa+log C] for weak acid strong base salt solution
Derivation for Strong base and Weak acid reaction.A- + H2O; HA + OH- Here HA is weak acid considered.As example, the salt CH3COONa. It ionises in water completely to give CH3COO- and Na+ ions. CH3COO- ions react with water to form a weak acid, CH3COOH and OH- ions.CH3COO- + H2O CH3COOH + OH-C(1-x) Cx CxThus, OH- ion concentration increases, the solution becomes alkaline.From law of mass action,Kh = [CH3COOH][OH-]/[CH3CO-] = (Cx×Cx)/C(1-x) = (Cx2)/(1-x) ) ...... (i)Other equations present in the solution are:CH3COOH CH3COO- + H+, Ka = [CH3COO-][H+]/[CH3COOH] ...... (ii)H2O H+ + OH-, Kw = [H+][OH-] ....... (iii)From eqs. (ii) and (iii),log [OH-] = log Kw - log Ka + log[salt]/[acid]-pOH = -pKw + pKa + log[salt]/[acid]pKw - pOH = pKa + log[salt]/[acid].pH=1/2[pKw+pKa+log C].Proved.
A- + H2O; HA + OH-
Here HA is weak acid considered.
As example, the salt CH3COONa. It ionises in water completely to give CH3COO- and Na+ ions. CH3COO- ions react with water to form a weak acid, CH3COOH and OH- ions.
CH3COO- + H2O CH3COOH + OH-
C(1-x) Cx Cx
Thus, OH- ion concentration increases, the solution becomes alkaline.
From law of mass action,
Kh = [CH3COOH][OH-]/[CH3CO-] = (Cx×Cx)/C(1-x) = (Cx2)/(1-x) ) ...... (i)
Other equations present in the solution are:
CH3COOH CH3COO- + H+,
Ka = [CH3COO-][H+]/[CH3COOH] ...... (ii)
H2O H+ + OH-,
Kw = [H+][OH-] ....... (iii)
From eqs. (ii) and (iii),
log [OH-] = log Kw - log Ka + log[salt]/[acid]
-pOH = -pKw + pKa + log[salt]/[acid]
pKw - pOH = pKa + log[salt]/[acid].
pH=1/2[pKw+pKa+log C].
Proved.
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