Question

Calculate the concentration of acid(or base) remaining in solution when 5.39mL of 6.93M HNO3 is added to 9.37mL of 8.37M NaOH. Note the final volume is the sum of two added volumes. which of the following statement is true for the solution after mixing?a) NaOH is in excess overHNO3b)HNO3 is in excess over NaOHc)HNO3 and NaOH are exactly balanced.What is the concentration of the excess NaOH (or HNO3) you indicated above?

Answer 1

Answer:

Concentration of remaining NaOH is 2.78 M.

The correct statement:

NaOH is in excess over $HNO_3$.

Explanation:

$Moles (n)=Molarity(M)\times Volume (L)$

Moles of nitric acid = n

Volume of nitric acid solution = 5.39 mL = 0.00539 L

Molarity of the nitric acid = 6.93 MM

$n=6.93 M\times 0.00539 L=0.03735 mol$

Moles of sodium hydroxide = n'

Volume of sodium hydroxide solution = 9.37 mL = 0.00937 L

Molarity of the sodium hydroxide = 8.37 M

$n'=8.37 M\times 0.00937 L=0.07843 mol$

$HNO_3+NaOH \rightarrow H_2O+NaNO_3$

According to reaction , 1 mole of nitric acid reacts with 1 mole of sodium hydroxide .

Then 0.03735 moles of nitric acid will neutralize 0.03735 moles of sodium hydroxide.

As we can see that sodium hydroxide is in excess amount. Hence excessive reagent and nitric acid is limiting reagent.

Moles of sodium hydroxide left = 0.07843 mol - 0.03735 mol = 0.04108 mol

Volume of the final solution  = 5.39 mL +9.37 ml = 14.76 ml = 0.01476 L

Concentration of remaining NaOH :

$=\frac{0.04108 mol}{0.01476 L}=2.78M$