As 50.0 mL of 0.10 M HCl is added to 100.0 mL of 0.10 M NaOH, what happens to the pH of the NaOH solution?
Sol :-
# pH of 100 mL of 0.10 M NaOH :
[NaOH] = [OH-] = 0.10 M
We know
pOH = - log [OH-]
pOH = - log 0.10 M
pOH = 1
As,
pH + pOH = 14
pH = 14 -pOH
pH = 14 - 1
pH = 13
So, pH of NaOH solution = 13
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# Addition of HCl :
(Eacid) Gram equivalents of Acid (HCl) = Normality x Volume in L
= Molarity x Valency factor of HCl x Volume of solution in L
= 0.10 M x 1 x 0.050 L
= 0.005
Similarly,
(Ebase)Gram equivalents of Base (NaOH) = Normality x Volume in L
= Molarity x Valency factor of NaOH x Volume of solution in L
= 0.10 M x 1 x 0.100 L
= 0.01
Because, gram equivalent of base are more, therefore pH of the resulting solution will be greater than 7.
[OH-] = Ebase-Eacid / VT
= 0.01 - 0.005 / 0.150 L
= 0.033
So,
pOH = - log 0.033
pOH = 1.48
pH = 14 - pOH
pH = 14 - 1.48
pH = 10.52
Hence, pH of NaOH solution decrease from 13 to 10.52 with the addition of HCl |
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