Question

Consider the neutralization reaction:

2 HNO3(aq) + Ba(OH)₂ (aq) → 2H₂O(1) + Ba(NO3)₂(aq)
A 0.120 L sample of an unknown HNO3 solution required 37.9 mL of 0.250 M Ba(OH), for complete neutralization. What is
the concentration of the HNO3 solution?

Answer 1

Step-by-step explanation:

First, we need to write a balanced chemical equation for the neutralization reaction:

2 HNO3(aq) + Ba(OH)2(aq) → 2 H2O(l) + Ba(NO3)2(aq)

From the balanced equation, we can see that the stoichiometric ratio of HNO3 to Ba(OH)2 is 2:1. This means that 2 moles of HNO3 react with 1 mole of Ba(OH)2.

Using the given information, we can calculate the number of moles of Ba(OH)2 that reacted:

moles of Ba(OH)2 = Molarity x Volume (in L)

moles of Ba(OH)2 = 0.250 M x (37.9/1000) L

moles of Ba(OH)2 = 0.009475 mol

Since the stoichiometric ratio of HNO3 to Ba(OH)2 is 2:1, the number of moles of HNO3 that reacted is twice the number of moles of Ba(OH)2:

moles of HNO3 = 2 x moles of Ba(OH)2

moles of HNO3 = 2 x 0.009475 mol

moles of HNO3 = 0.01895 mol

Finally, we can calculate the concentration of the HNO3 solution:

concentration of HNO3 = moles of HNO3 / volume of HNO3 solution (in L)

concentration of HNO3 = 0.01895 mol / 0.120 L

concentration of HNO3 = 0.158 mol/L

Therefore, the concentration of the HNO3 solution is 0.158 mol/L.