Respuesta :
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 = [tex]\frac{15.0 g}{40 g/mol}=0.375 mol[/tex]
Molarity of the nitric acid solution = 0.250 M
Volume of the nitric solution = 0.150 L
Moles of nitric acid = n
[tex]Molarity=\frac{Moles}{Volume(L)}[/tex]
[tex]n=0.250 M\times 0.150 L=0.0375 mol[/tex]
[tex]NaOH+HNO_3\rightarrow NaNO_3+H_2O[/tex]
According to reaction, 1 mole of nitric acid recats with 1 mole of NaOH, then 0.0375 moles of nitric acid will react with :
[tex]\frac{1}{1}\times 0.0375 mol[/tex] of NaOH
Moles of NaOH left unreacted in the solution =
= 0.375 mol - 0.0375 mol = 0.3375 mol
[tex]NaOH(aq)\rightarrow Na^+(aq)+OH^-(aq)[/tex]
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 :
[tex]1\times 0.3375 moles=0.3375 mol[/tex] of hydroxide ion
The molarity of hydroxide ion in solution ;
[tex]=\frac{0.3375 mol}{0.150 L}=2.25 M[/tex]
2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.
