Question

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!!

Answer 1

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

Answer 2

because at end point the solution is neutral, so the # of moles are the same? No ...so what do I do differently??

Answer 3

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.