Question

A solution is prepared from 0.02 mol of MgCl2 and 0.004 mol of NaOH in 1L of water. What is the PH of the solution? the Ksp of Mg(OH)2 is 6 x10^-12

Answer 1

the reaction is

MgCl2 + 2NaOH ---> Mg(OH)2 (s) + 2NaCl

we can see that

moles of MgCl2 reacted = 0.5 x moles of NaOH

moles of MgCl2 reacted = 0.5 x 0.004 = 2 x 10-3

now

moles of Mg(OH)2 formed = moles of MgCl2 reacted = 2 x 10-3

now

moles of MgCl2 left = 0.02 - ( 2 x 10-3) = 0.018

now

concentration = moles / volume (L)

so

conc of MgCl2 = 0.018 / 1 = 0.018

MgCl2 ---> Mg+2 + 2Cl-

we know that

MgCl2 is soluble

so

[Mg+2] = [MgCl2] = 0.018

now

Mg(OH)2 is not soluble

we get

Mg(OH)2 (s)---> Mg+2 + 2OH-

Ksp = [Mg+2] [OH-]^2

6 x 10-12 = [0.018] [OH-]^2

[OH-]= 1.826 x 10-5

now

pOH = -log [OH-]

pOH = -log 1.826 x 10-5

pOH = 4.74

now

pH = 14 - pOH

so

pH = 14 - 4.74

pH = 9.26

so

pH of the solution is 9.26