In: Chemistry
A 0.120-L sample of an unknown HNO3 solution required 29.1 mL of 0.200 M Ba(OH)2 for complete neutralization. What was the concentration of the HNO3 solution?
Given Molarity of Ba(OH)2 = 0.200 M
Volume of Ba(OH)2 = 29.1 mL = 0.0291 L Sice 1 L = 1000 mL
So number of moles of Ba(OH)2 is n = Molarity x volume in L
= 0.200 M x 0.0291 L
= 0.00582 mol
Ba(OH)2 + 2HNO3 Ba(NO3)2 + 2H2O
According to the balanced equation ,
1 mole of Ba(OH)2 reacts with 2 moles of HNO3
0.00582 mole of Ba(OH)2 reacts with 2x0.00582 = 0.01164 moles of HNO3
So molarity of HNO3 is ,
So the concentration of HNO3 is 0.097 M