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
The pH at the equivalence point when a solution of 0.01 M-CH3COOH is titrated with a solution of 0.01 M-NaOH, is (pKa of...The solubility product of AgC2O4 at 25°C is 2.3 × 10–11 M3. A solution of K2C2O4 containing 0.15 moles in 500 ml water i...On adding 0.1 M solution each of [Ag+], [Ba2+], [Ca2+] in a Na2So4 solution, species first precipited is...50 mL of 0.1 M NaOH is added to 60 mL of 0.15 M H3PO4 solution (K1, K2 and K3 for H3PO4 are 10-3, 10-8 and 10-13 respect...Let the solubility of an aqueous solution of Mg(OH)2 be x then its ksp is...