A 0.100 L sample of an unknown HNO3 solution required 51.3 mL of 0.250 M Ba(OH)2 for complete neutralization. What is the concentration of the HNO3 solution

0.257M HNO3

Explanation:

The reaction of HNO3 with Ba(OH)2 is:

2 HNO₃ + Ba(OH)₂ → 2 H₂O + Ba(NO₃)₂

That means 2 moles of HNO3 reacts with 1 mole of Ba(OH)2.

51,3mL of 0.250M Ba(OH)2 are:

0.0513L ₓ (0.250mol / L) = 0.0128moles

Moles of HNO3 are:

0.0128moles Ba(OH)2 ₓ (2 mol HNO3 / 1 mol Ba(OH)2) = 0.0257mol HNO3

As these moles are in 0.100L:

0.0257mol HNO3 / 0.100L = 0.257M HNO3

I hope it helps!