Question

3. Determine the pH (to two decimal places) of the solution that is produced by mixing 5.05 mL of 7.91×10^-5 M HI with 7.73 mL of 8.38×10^-5 M Mg(OH)₂.

Answer 1

2 HI + Mg(OH)2 ------------> MgI₂ + 2H₂O

Mg(OH)₂:

Molarity = 8.38 x 10^-5 M

Normality = 2 x molarity = 2 x 8.38 x 10^-5 = 16.76 x 10^-5 N

volume = 7.73 mL = 0.00773 L

Equivalents of Mg(OH)2 = Normality x volume = 16.76 x 10^-5 N X 0.00773 L= 0.13 x 10^-5

HI

Molarity = 7.91×10^-5 M

Normality = 7.91×10^-5 M

Volume = 5.05 mL = 0.00505 L

Equivalents of HI = Normality x volume = 7.91×10^-5 M X 0.00505 L = 0.04 x 10^-5

But Equivalents of Mg(OH)2 = 0.13 x 10^-5

Therefore, concentration of Equivalents of Mg(OH)2 is more than concentration of Hl.

Hence at equivalence point, concentration of OH- is higher than concentration of H+ ions.

[OH-] = Equivalents of Mg(OH)2 - Equivalents of HI / total volume (volume of Mg(OH)2+ Hl)

= (0.13 x 10^-5 - 0.04 x 10^-5) /(0.00773 L+ 0.00505 L)

= 0.0000704

[OH-] = 0.0000704 N

pOH = -log[OH-]

= -log(0.0000704 )

= 4.15

pOH = 4.15

Hence, pH = 14-pOH = 14 - 4.15 = 9.85

Therefore, pH = 9.85