Question

How many milliliters of 0.534 M Ba(OH)2 solution are needed to completely neutralize 34.5 mL of 0.220 M HBr solution? The reaction is

Ba(OH)2(aq) + 2HBr(aq) → BaBr2(aq) + 2H2O(l)

I did

0.0345 * 0.220 = 0.00759
Moles of HCl =2*0.00759= 0.01518

0.01518*1000= 15.18 mL Answer

It was wrong. Can anybody help me please.

Answer 1

You are right with 0.0759. That is mols HBr.
Then mols Ba(OH)2 is 1/2 that and not 2x that.

Finally, the third is an error also. It should be M Ba(OH)2 = mols Ba(OH)2/L Ba(OH)2. You know mols and M, solve L and convert to mL.

Answer 2

I still have problem on how can you get mols Ba(OH)2 from. Can you explain please

Answer 3

Ba(OH)2 is 1/2 How can you get 1/2 from.

Answer 4

so i have to do

1/2*0.00759=0.003795

= 3.795 mL is the answer?

Answer 5

Ba(OH)2(aq) + 2HBr(aq) → BaBr2(aq) + 2H2O(l)
By the way, I see I made a typo late last night. I said your 0.0759 mols HBr was right; I should have said your 0.00759 mols HBr.
Here is how you get the 1/2. Use the coefficients in the balanced equation.
0.00759 mols HBr x (1 mol Ba(OH)2/2 mols HBr) = 0.00759 x 1/2 = 0.003795 mols Ba(OH)2.

Then M Ba(OH)2 = mols Ba(OH)2/L Ba(OH)2.
M Ba(OH)2 is 0.534 from the problem.
mols Ba(OH)2 = 0.003795 from above.
Solve for L. L = mols/M = 0.003795/0.534 = 0.007106L which is 7.106 mL and that would round to 7.11 mL to 3 s.f..

Answer 6

oh i got it... thank you so much