Question

Suppose a solution is prepared by dissolving 15.0 g NaOH in 0.150 L of 0.250 M nitric acid. What is the final concentration of OH− ions in the solution after the reaction has gone to completion. Assume that there is no volume change when adding the grams of NaOH

Answer 1

Answer:

2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.

Explanation:

Moles of NaOH = $\frac{15.0 g}{40 g/mol}=0.375 mol$

Molarity of the nitric acid solution = 0.250 M

Volume of the nitric solution = 0.150 L

Moles of nitric acid = n

$Molarity=\frac{Moles}{Volume(L)}$

$n=0.250 M\times 0.150 L=0.0375 mol$

$NaOH+HNO_3\rightarrow NaNO_3+H_2O$

According to reaction, 1 mole of nitric acid recats with 1 mole of NaOH, then 0.0375 moles of nitric acid will react with :

$\frac{1}{1}\times 0.0375 mol$ of NaOH

Moles of NaOH left unreacted  in the solution =

= 0.375 mol - 0.0375 mol = 0.3375 mol

$NaOH(aq)\rightarrow Na^+(aq)+OH^-(aq)$

1 mole of sodium hydroxide gives 1 mol of sodium ions and 1 mole of hydroxide ions.

Then 0.3375 moles of NaOH will give :

$1\times 0.3375 moles=0.3375 mol$ of hydroxide ion

The molarity of hydroxide ion in solution ;

$=\frac{0.3375 mol}{0.150 L}=2.25 M$

2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.