Hydrochloric acid (68.0 mL of 0.037 M) is added to 249.0 mL of 0.062 M Ba(OH)2 solution. What excess ions are left in the solution?

To solve this problem, we need to first find the balanced chemical equation for the reaction between hydrochloric acid (HCl) and barium hydroxide (Ba(OH)2):

2HCl + Ba(OH)2 → BaCl2 + 2H2O

From the equation, we can see that for every 2 moles of HCl, we need 1 mole of Ba(OH)2 to react completely.

Step 1: Calculate the moles of HCl and Ba(OH)2

Moles of HCl = (Volume of HCl solution in L) x (Concentration of HCl in M)
= 0.068 L x 0.037 M
= 0.002516 moles

Moles of Ba(OH)2 = (Volume of Ba(OH)2 solution in L) x (Concentration of Ba(OH)2 in M)
= 0.249 L x 0.062 M
= 0.015438 moles

Step 2: Determine the limiting reactant
The limiting reactant is the one that is completely consumed and determines the amount of product formed. In this case, we can see that 2 moles of HCl react with 1 mole of Ba(OH)2. Therefore, the limiting reactant is HCl because we have fewer moles of HCl than Ba(OH)2.

Step 3: Calculate the moles of product formed
Since HCl is the limiting reactant, we can use the ratio from the balanced equation to determine the moles of BaCl2 formed.

Moles of BaCl2 = (moles of HCl) x (1 mole BaCl2/2 moles HCl)
= 0.002516 moles x (1/2)
= 0.001258 moles

Step 4: Calculate the moles of excess ions
To find the moles of excess ions left in the solution, we need to subtract the moles of product formed from the initial moles of Ba(OH)2.

Moles of excess ions = Moles of Ba(OH)2 - Moles of BaCl2
= 0.015438 moles - 0.001258 moles
= 0.01418 moles

Step 5: Convert moles of excess ions to volume
To convert moles of excess ions to volume, we need to use the concentration of Ba(OH)2 solution.

Volume of excess ions = (moles of excess ions) / (Concentration of Ba(OH)2 in M)
= 0.01418 moles / 0.062 M
= 0.229 mL

So, there are 0.229 mL of excess ions left in the solution after the reaction between HCl and Ba(OH)2.