How many mL of 0.100M HCl are needed to neutralize 49.0mL of 0.104M Ba(OH)2 solution?
Ba(OH)2 + 2HCl ==> 2H2O + BaCl2
mols Ba(OH)2 = M x L = ?
Convert mols Ba(OH)2 to mols HCl using the coefficients in the balanced equation. mols HCl = twice mols Ba(OH)2.
M HCl = mols HCl/L HCl.
Solve for L and convert to mL.
To find the answer, we can use the concept of stoichiometry and the balanced chemical equation for the reaction between HCl and Ba(OH)2.
The balanced chemical equation for the reaction is as follows:
2 HCl + Ba(OH)2 -> BaCl2 + 2 H2O
From the equation, we can see that the ratio between HCl and Ba(OH)2 is 2:1. This means that for every 2 moles of HCl, we need 1 mole of Ba(OH)2.
First, let's find the number of moles of Ba(OH)2 in 49.0 mL of 0.104M Ba(OH)2 solution:
moles of Ba(OH)2 = volume (in L) x concentration (in mol/L)
= 49.0 mL x (1 L / 1000 mL) x 0.104 mol/L
= 0.005096 moles
Since the ratio of HCl to Ba(OH)2 is 2:1, we need half as many moles of HCl as moles of Ba(OH)2.
moles of HCl = 0.005096 moles / 2
= 0.002548 moles
Now, we can determine the volume of 0.100M HCl solution needed to neutralize this amount of moles:
volume (in L) = moles / concentration
= 0.002548 moles / 0.100 mol/L
= 0.02548 L
Finally, convert the volume to milliliters:
volume (in mL) = 0.02548 L x (1000 mL / 1 L)
= 25.48 mL
Therefore, 25.48 mL of 0.100M HCl solution is needed to neutralize 49.0 mL of 0.104M Ba(OH)2 solution.