Question

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?

Answer 1

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 :

(E_acid) 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,

(E_base)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^-] = E_base-E_acid / V_T

= 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 ​​​​​​​