Question #036ed

2 Answers
Mar 31, 2017

Answer:

The mixture would be water and sodium chloride at a concentration of .1 molar.

Explanation:

Sodium Hydroxide would neutralize the Hydrochloric acid.

# Na^+1 + OH^-1 + H^+1 + Cl^-1 === H_2O + NaCl #

Since the molar concentration of the Sodium Hydroxide would be .2 molar, the concentration of the Sodium ion would be .2 molar.
Since the molar concentration of the Hydrogen Chloride ( Hydrochloric acid) would be .2 molar the concentration of the Chlorine ion would be .2 molar.

These concentrations would be before the mixture. Since the mixture would have twice the volume of the initial solutions and the moles of the Sodium Chloride doesn't change the concentration of the solution of the mixture is half

# .2mol/( 2 liters) = .1 mol/(liter)#

The molar concentration of the salt water is .1 M

Apr 18, 2017

Answer:

The mixed solution will consist of 500 mL of 0.2 mol/L #"H"_2"SO"_4# and 0.05 mol/L #"Na"_2"SO"_4#.

Explanation:

#"Initial moles of H"_2"SO"_4 = 0.250 color(red)(cancel(color(black)("L"))) × "0.5 mol"/(1 color(red)(cancel(color(black)("L")))) = "0.125 mol"#

#"Initial moles of NaOH" = 0.250 color(red)(cancel(color(black)("L"))) × "0.2 mol"/(1 color(red)(cancel(color(black)("L")))) = "0.05 mol"#

The equation for the reaction is

#"H"_2"SO"_4 + "2NaOH" → "Na"_2"SO"_4 + "2H"_2"O"#

This looks like a limiting reactant problem, so we calculate the amount of #"Na"_2"SO"_4# that we can get from each reactant.

Identify the limiting reactant

From #"H"_2"SO"_4#:

#"Moles of Na"_2"SO"_4 = 0.125 color(red)(cancel(color(black)("mol H"_2"SO"_4))) × ("1 mol Na"_2"SO"_4)/(1 color(red)(cancel(color(black)("mol H"_2"SO"_4)))) = "0.125 mol Na"_2"SO"_4#

From #"NaOH"#:

#"Moles of Na"_2"SO"_4 = 0.05color(red)(cancel(color(black)("mol NaOH")))× ("1 mol Na"_2"SO"_4)/(2 color(red)(cancel(color(black)("mol NaOH")))) = "0.025 mol Na"_2"SO"_4#

#"NaOH"# gives the fewest moles of #"Na"_2"SO"_4#, so #"NaOH"# is the limiting reactant.

Calculate the moles of #"H"_2"SO"_4# reacted

#"Moles of H"_2"SO"_4color(white)(l) "reacted" = 0.05 color(red)(cancel(color(black)("mol NaOH"))) × ("1 mol H"_2"SO"_4)/(2 color(red)(cancel(color(black)("mol NaOH")))) = "0.025 mol H"_2"SO"_4#

Calculate the moles of #"H"_2"SO"_4# remaining

#"Moles of H"_2"SO"_4color(white)(l) "remaining" = "0.125 mol - 0.025 mol" = "0.1 mol"#

Calculate the final concentrations of all substances

At the end of the reaction, we have 0.1 mol #"H"_2"SO"_4# and 0.025 mol #"Na"_2"SO"_4#
in 500 mL solution.

#["H"_2"SO"_4] = "0.1 mol"/"0.5 L" = "0.2 mol/L"#

#["Na"_2"SO"_4] = "0.025 mol"/"0.5 L" = "0.05 mol/L"#