When excess magnesium nitrate was added to 175 mL of a solution of sodium hydroxide, a solid precipitate formed.

Question

When excess magnesium nitrate was added to 175 mL of a solution of sodium hydroxide, a solid precipitate formed. The dry precipitate had a mass of 11.43 g. What was the concentration of hydroxide ions in the solution of sodium hydroxide?

Answer 1

First write the chemical equation:

 

Mg(NO3)2 + NaOH → Mg(OH)2 + NaNO3

 

Balance:

 

Mg(NO3)2 + 2NaOH → Mg(OH)2 + 2NaNO3

 

Compute moles of Mg(OH)2:

 

11.43 g Mg(OH)2 x 1 mole Mg(OH)2/58.32 g Mg(OH)2 = 0.196 moles of Mg(OH)2

 

That is 0. moles of OH-, because there are two OH- in each formula unit of Mg(OH)2

 

So, you know that .39 moles of OH- from the NaOH solution was used to form the Mg(OH)2 that precipitated.

 

.39 moles of OH-/.175L of NaOH = 2.23 Mole of NaOH/L