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.

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.

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

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

so i have to do

1/2*0.00759=0.003795

= 3.795 mL is the answer?

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..

oh i got it... thank you so much

To find the correct answer, you need to use the concept of stoichiometry to determine the molar ratio between Ba(OH)2 and HBr in the balanced chemical equation.

First, let's find the moles of HBr in the 34.5 mL of 0.220 M HBr solution:

Moles of HBr = Volume (L) x Concentration (M)

Volume = 34.5 mL = 0.0345 L
Concentration = 0.220 M

Moles of HBr = 0.0345 L x 0.220 M = 0.00759 mol

Next, we use the balanced chemical equation to determine the molar ratio between Ba(OH)2 and HBr. From the balanced equation, we know that 1 mole of Ba(OH)2 reacts with 2 moles of HBr.

Therefore, using the ratio:

Moles of Ba(OH)2 = (Moles of HBr) / 2

Moles of Ba(OH)2 = 0.00759 mol / 2 = 0.0038 mol

Now that we have the moles of Ba(OH)2, we can determine the volume of the 0.534 M Ba(OH)2 solution needed to completely neutralize it. We use the same formula as before:

Volume (L) = Moles / Concentration

Moles = 0.0038 mol
Concentration = 0.534 M

Volume (L) = 0.0038 mol / 0.534 M = 0.0071 L

Finally, convert the volume from liters to milliliters:

Volume (mL) = Volume (L) x 1000

Volume (mL) = 0.0071 L x 1000 = 7.1 mL

Therefore, you need 7.1 mL of a 0.534 M Ba(OH)2 solution to completely neutralize 34.5 mL of a 0.220 M HBr solution.