Question

Determine the pH of a solution when 22.7 mL of 0.028 M HNO3 is mixed with 19.6 mL of

A) 0.0120 M NaOH.
B) distilled water
C) 0.0080 M HCl
D) 0.144 M KOH

Answer 1

Solution :

Lets show reaction of HNO3 with NaOH

Formula used :

Mol = Molarity x volume in L

And

Molarity = mol / volume in L

Lets calucate moles of HNO3

n HNO3 = 0.028 M x 0.0227 L

=6.36 E-4 mol

n NaOH = 0.0120 x 0.0196 L=2.35 E-4

Both have 1:1 mol ratio

So limiting reactant is NaOH

Calculation of moles of excess acid

n HNO3 = 6.36 E-4 – 2.35 E-4 = 4.01 E-4

since given acid is strong acid so it dissociates completely in the solution

pH = [HNO3]

[HNO3]= 4.01 E-4 / (0.0227 + 0.0196 )L

=0.00948 M

pH = - log (0.00948)

=2.023

B) Distilled water

Distilled water only dilutes the solution and no any reaction is there since acid is strong.

Lets calculate final concentration of acid

M2 = M1V1/V2 = 0.028 M x 0.0227 L / ( 0.0227 L+0.0196) L =0.01503 M

pH = -log (0.01503)

= 1.82

C) 0.0080 M HCl

Moles of HCl = 0.0080 M x 0.0196 L =1.57 E-4 mol

Total moles = n HCl + n HNO3 = 0.00082 mol

Total [H+] = 0.00082 mol / (( 0.0227 L+0.0196 ) L

= 0.0194 M

pH = 1.71

D : 0.144 M KOH

This also reacts same as like NaOH \

n KOH = 0.144 M x 0.0196 L = 0.002822 mol

This is excess

Excess moles of KOH = (0.002800 – 6.36 E-4 ) mol

= 0.002186 mol

[KOH]= 0.002186 / (( 0.0227 L+0.0196) L

= 0.051688 M

pH = 14 – (-log (OH-) = 12.71