Question

when 500 ml of 0.10 M NaOH solution (containing Na+ and OH-ions) is mixed with 500 ml of 0.10 M Mg(NO3)2 solution containing Mg2+ and NO3- ions, a precipitate of solid Mg(OH)2 forms and the resulting aqueous solution has ph=9.43. Based on the information, determine the value of ksp for Mg(OH)2. Show your reasoning clearly

Answer 1

2NaOH(aq) + Mg(NO3)2(aq) ------------> Mg(OH)2(s) + 2NaNO3 (a)

500x0.1 500x0.1

= 50 50 0 0 initial mmoles

0 25 25 - after reation

That is all NaOH is reacted and the [OH-] in the solution is only from the Mg(OH)2 dissociation

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

s s 2s

Ksp = [Mg+2] [OH-]²

Now the pH = 9.43 after precipitation

so pOH = 14-9.43 = 4.57 and [Oh-] = - antilog 4.57

Thus [OH-] = 3.715x 10 ^-5 M

Hence [Mg+2] = 1/2 [OH-] = 3.715x 10 ^-5 M/2

thus Ksp of Mg(OH)2 = 3.715x 10 ^-5 M/2 (3.715x 10 ^-5 M)²

= 2.56 x 10^-14