Question

When 220 mL of 1.50x10^-4 M hydrochloric acid is added to 135 mL of 1.75x10^-4 M Mg(OH)2, the resulting solution will be

Answer 1

The given reaction of HCl and Mg(OH)2 is as follows:

Mg(OH)2(s) + 2HCl(aq) → MgCl2(aq) +2H2O(l)

Number of moles = volume in L * molarity

HCl moles: 220/1000*1.50*10^-4

= 3.3*10^-5

Mg(OH)2: = 135/1000*1.75*10^-4

= 2.3625*10^-5 Moles

Here HCl is limiting agent

The limiting agent has due to following properties:

1. It completely reacted in the reaction.
2. It determines the amount of the product in mole.

Moles of Mg(OH)2 which area reacted with HCl:

3.3*10^-5 HCl *1 mole Mg(OH)2 / 2 Mole HCl

= 1.65*10^-5 Mg(OH)2

Reaming mole Mg(OH)2= 2.3625*10^-5 Moles -1.65*10^-5 Mg(OH)2

= 7.125*10^-6 Mg (OH)2 OR 1.425*10^-6 mole OH-

Molarity = number of moles/ total volume in L

= 1.425*10^-6 mole OH-/0.355

= 4.02*10^-6

pOH = - log [OH-]

= - log 4.02*10^-6

=5.40

pH = 14- pOH

= 14-5.40

= 8.60 thus it is basic solution