This document calculates the concentrations of ions remaining after mixing solutions containing magnesium nitrate and sodium hydroxide. It first calculates the molarities of the reactants and uses the solubility product constant to determine the solubility of magnesium hydroxide. It then determines the final concentrations of magnesium, nitrate, sodium, and hydroxide ions are 1.65x10^-4 M, 0.4 M, 0.563 M, and 3.3x10^-4 M, respectively.

1. value 0.41 points 1 out of 3 attempt Be sure to answer all parts. A 22.59-mL solution containing
1640 g Mg(NO,), is mixed with a 32.12-mL solution containing 1.231 g NaOH. Calculate the
concentrations of the ions remaining in solution after the reaction is complete. Assume volumes
are additive. If a species fully precipitates, type o. M NO, M Na ???- References Multipart
Answer 3
Solution
no of mol of Mg(NO3)2 = 1.64/148.3 = 0.011 mol
concentration of Mg(NO3)2 = (0.011/(22.59+32.12))*1000 = 0.2 M
NO of mol of NaOH = 1.231/40 = 0.0308 mol
concentration of NaOH in the mixture = (0.0308/ (22.59+32.12))*1000
= 0.563 M
Mg(OH)2(s) -----> Mg2+(aq) + 2OH-(aq)
ksp of Mg(OH)2 = s*(2s)^2 = 1.8*10^-11
s = solubility of Mg(oH)2 = 1.65*10^-4 M
finally
[Mg2+] = S = 1.65*10^-4 M
[NO3-] = 0.2*2 = 0.4 M
[Na+] = 0.563 M
[OH-] = 2S = 1.65*10^-4*2 = 3.3*10^-4 M