Question

In a titration, 21.5 mL of 0.00369 M Ba(OH)2 neutralized 12.1 mL of HCl solution. What is the molarity of the HCL solution?
( I used the titration formula and plugged in the given values. But it's wrong. I know we don't have to divide by any number to the molarity of HCL because there is only one H) pls. help thank you

Answer 1

there are two OH in the barium solution

21.5 * 0.00369 M * 2 = 12.1 * x M

the answer should have three significant figures ... no more , no less

Answer 2

Ba (OH)2 + 2HCl --> BaCl2 + 2H2O
how many liters of fluid? (21.6+12.1)10^-3 = 33.7 *10^-3 L
mols of Ba(OH)2 = .00369 mols/L * 21.5 *10^-3 L = .0793 *10^-3mols
so need 2 * .0793 *10^-3 = 0.159 * 10^-3 mols HCL
0.159 * 10^-3 mols/ 12.1*10^-3 L
= 0.0131 M HCL