The titration of 25.00mL of Ba(OH)2 solution with an aqueous solution of 0.500 M HCl reaches an end point when 30.20 mL of the HCl solution is added. Determine the concentration of M of the Ba(OH)2 solution.
Soo 25.00mL=0.025L
30.20mL= 0.0302L
moles= MxL (0.500M)(0.0302L)=0.5302 moles of HCl and since the moles are the same at the end point than 0.5302 is moles of Ba(OH)2 too?
M=moles Ba(OH)2/initial L 0.5302/0.025L= 21.2M is my answer
but the book says the answer is 0.302M
what have I done wrong?? Thanks!!
Why do you think the moles HCl and Ba(OH)2 are the same? Did you write the equation?
2HCl + Ba(OH)2 ==> BaCl2 + 2H2O
because at end point the solution is neutral, so the # of moles are the same? No ...so what do I do differently??
The solution may be neutral BUT the two reagents do not react 1:1 and we must correct for that (and the moles are not the same).
2HCl + Ba(OH)2 ==> BaCl2 + 2H2O
moles HCl = M x L = 0.500 x 0.03020 = 0.01510 (I don't know where your number came from; probably punched into the calculator wrong since the numbers are correct.).
Now convert moles HCl to moles Ba(OH)2.
moles Ba(OH)2 = moles HCl x (1 mole Ba(OH)2/2 moles HCl) = 0.01510 x (1/2) = 0.00755 moles Ba(OH)2.
M Ba(OH)2 = mols/L = 0.00755/0.025 = 0.3020 which rounds to 0.302 to three s.f.
To determine the concentration of the Ba(OH)2 solution, we need to use the balanced chemical equation for the reaction between Ba(OH)2 and HCl:
Ba(OH)2 + 2HCl -> BaCl2 + 2H2O
From the equation, we can see that 1 mole of Ba(OH)2 reacts with 2 moles of HCl.
First, calculate the number of moles of HCl used in the reaction:
Moles of HCl = concentration of HCl × volume of HCl solution used
= 0.500 M × 0.0302 L
= 0.0151 moles
Since the stoichiometry of the reaction is 1:2 (Ba(OH)2:HCl), the number of moles of Ba(OH)2 used is also 0.0151 moles.
Now, to find the concentration of the Ba(OH)2 solution, divide the moles of Ba(OH)2 by the volume of the solution used (in liters):
Concentration (M) = Moles of Ba(OH)2 / Volume of Ba(OH)2 solution used
= 0.0151 moles / 0.025 L
= 0.604 M
So, the concentration of the Ba(OH)2 solution is approximately 0.604 M, not 21.2 M or 0.302 M.
It seems there was an error in your calculations. The mistake may have come from using the initial volume of the Ba(OH)2 solution (25.00 mL) instead of the volume of the HCl solution used (30.20 mL) in the final concentration calculation. Please double-check your calculations following the steps provided above to obtain the correct answer.