Ionic EquilibriumHard
Question
The concentration of hydroxyl ion in a solution left after mixing 100 mL of 0.1 M MgCI2 and 100 mL of 0.2 M NaOH (KSP of Mg(OH)2 = 1.2 × 10-11):
Options
A.2.88 × 10-3
B.2.88 × 10-2
C.2.88 × 10-4
D.2.88 × 10-5
Solution
MgCl2 + 2NaOH → Mg(OH)2 + 2NaCl
mm before 10 20 0 0
reaction 0 0 10 20
Thus, 10 m mole of Mg(OH)2 are formed. The product of [Mg2+][OH-]2 is therefore
= 5 √ 10-4 which is more than KSP of Mg(OH)2. Now solubility(s) of Mg(OH)2 can be derived by
KSP = 4s3
∴ s =
= 1.4 × 10-4 M
∴ [OH-] = 2s = 2.88 × 10-4 M
mm before 10 20 0 0
reaction 0 0 10 20
Thus, 10 m mole of Mg(OH)2 are formed. The product of [Mg2+][OH-]2 is therefore
KSP = 4s3
∴ s =
∴ [OH-] = 2s = 2.88 × 10-4 M
Create a free account to view solution
View Solution FreeMore Ionic Equilibrium Questions
Buffer solutions have constant acidity and alkalinity because...For which of the following solutions must we consider the ionisation of water when calculating the pH or pOH :-...Which one of the following substances has the highest proton affinity?...An example of a reversible reaction is :...Which is most stable ?...