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
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
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
To find the molarity of the HCl solution, we can start by writing a balanced chemical equation for the neutralization reaction between Ba(OH)2 and HCl:
Ba(OH)2 + 2HCl -> BaCl2 + 2H2O
From the balanced equation, you can see that 1 mole of Ba(OH)2 reacts with 2 moles of HCl.
Given that 21.5 mL of 0.00369 M Ba(OH)2 reacts with 12.1 mL of HCl solution, we want to find the molarity of HCl.
To do this, we need to use the concept of stoichiometry, which relates the number of moles of a reactant to the number of moles of another reactant based on the balanced equation.
First, let's find the number of moles of Ba(OH)2 used in the reaction:
Moles of Ba(OH)2 = volume (in liters) * concentration (in moles per liter)
Moles of Ba(OH)2 = 0.0215 L * 0.00369 mol/L
Moles of Ba(OH)2 = 7.9135 x 10^-5 mol
Since the stoichiometry tells us that 1 mole of Ba(OH)2 reacts with 2 moles of HCl, the number of moles of HCl used in the reaction is twice that of Ba(OH)2:
Moles of HCl = 2 * Moles of Ba(OH)2
Moles of HCl = 2 * 7.9135 x 10^-5 mol
Moles of HCl = 1.5827 x 10^-4 mol
Next, we need to find the molarity of HCl. Since we have the volume of HCl in liters (12.1 mL = 0.0121 L), we can use the formula:
Molarity (M) = Moles of solute / Volume of solution (in liters)
Molarity of HCl = (1.5827 x 10^-4 mol) / 0.0121 L
Molarity of HCl = 0.0131 M
Therefore, the molarity of the HCl solution is 0.0131 M.