Question

100ml of 0.1M H2SO4 & 900ml of 0.1M NAOH find its nature and normality​

Answer 1

Explanation:

A mixture containing 100 ml of 0.1 M acetic acid and 50 ml of 0.1 M NaOH is acidic buffer soluion. Its pH is given by the expression pH=pKa+logacidsalt.Substitute values in the above expression.5=pKa+log0.1×0.050.1×0.05 or pKa=5.When 100 ml of 0.05 M NaOH  is added, the acid is completely neutralized and the solution contains salt sodium acetate.The expression for the hydrogen ion concentration of the salt of weak acid and strong base is pH=21(pKw+pKa+logc).Substitute values in the above expression.pH=21(14+5+log(0.250.1))=8.8.

Hence, the change in pH is ΔpH=8.8−5=3.8.

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