Hydrochloric acid(75.0 ml of 0.250 M) is added to 225.0 ml of 0.0550 M Barium hydroxide solution. What is the concentration of the excess H+ or OH- ions left in this solution?
2HCl + Ba(OH)2 ==> 2H2O + BaCl2
mols HCl = M x L = ?
mols Ba(OH)2 = M x L = ?
Determine the limiting reagent and go from there. Post your work if you get stuck and I can help you through.
mols HCl= 0.01875
mols Ba(OH)2 = 0.012375
Well, it seems like we've got a bit of a chemical party going on here! Let's see what's happening.
First, we need to find out how many moles of each compound we have. By using the formula C = n/V (where C is concentration, n is moles, and V is volume), we find that we have 0.01875 moles of HCl (0.250 M * 0.075 L) and 0.012375 moles of Ba(OH)2 (0.0550 M * 0.225 L).
Now, let's use the balanced equation to determine what's going on here. The reaction between HCl and Ba(OH)2 goes like this: HCl + Ba(OH)2 -> BaCl2 + H2O.
We can tell that for every one mole of HCl, we need two moles of Ba(OH)2 to react completely. Since we have fewer moles of Ba(OH)2 (0.012375) compared to HCl (0.01875), we know that the Ba(OH)2 is the limiting reactant.
So, all of the Ba(OH)2 reacted, and we have some amount of excess HCl. A mole of HCl yields one mole of H+ ions in solution, so we have the same number of moles of the excess H+ ions (0.012375 moles) remaining in the solution.
In terms of concentration, we need to calculate it by dividing the number of moles of excess H+ ions by the total volume of the solution (75.0 mL + 225.0 mL = 300.0 mL or 0.3 L).
Concentration = moles/volume = 0.012375 moles / 0.3 L ≈ 0.041 M.
So, the concentration of the excess H+ ions left in the solution is approximately 0.041 M. Keep calm and carry on balancing those chemical equations!
To find the concentration of the excess H+ or OH- ions left in the solution, we need to determine which reagent is the limiting reactant and calculate the amount of excess H+ or OH- ions produced.
First, let's determine the limiting reactant. We can do this by comparing the number of moles of each reactant.
Given:
Volume of HCl solution (V1) = 75.0 ml
Concentration of HCl solution (C1) = 0.250 M
Volume of Ba(OH)2 solution (V2) = 225.0 ml
Concentration of Ba(OH)2 solution (C2) = 0.0550 M
To convert volumes to liters, divide by 1000:
V1 = 75.0 ml / 1000 = 0.075 L
V2 = 225.0 ml / 1000 = 0.225 L
Using the molarity formula:
Molarity (M) = moles (n) / volume (V)
For HCl:
n1 = C1 * V1 = 0.250 M * 0.075 L = 0.01875 moles
For Ba(OH)2:
n2 = C2 * V2 = 0.0550 M * 0.225 L = 0.012375 moles
From the balanced chemical equation:
Ba(OH)2 + 2HCl -> 2H2O + BaCl2
We can see that 1 mole of Ba(OH)2 reacts with 2 moles of HCl.
Since the molar ratio is 1:2, the moles of HCl needed to react with all the Ba(OH)2 is:
(0.012375 moles of Ba(OH)2) * (2 moles of HCl / 1 mole of Ba(OH)2) = 0.02475 moles of HCl
Therefore, since we have only 0.01875 moles of HCl, it is the limiting reactant. This means that all of the HCl will react, and we will have an excess of OH- ions.
To calculate the concentration of the excess OH- ions:
Since 1 mole of Ba(OH)2 produces 2 moles of OH- ions, we can calculate the moles of OH- ions produced:
(0.012375 moles of Ba(OH)2) * (2 moles of OH- ions / 1 mole of Ba(OH)2) = 0.02475 moles of OH- ions
Now we need to calculate the volume of the final solution, which is the sum of the initial volumes of HCl and Ba(OH)2 solutions:
V_total = V1 + V2 = 0.075 L + 0.225 L = 0.3 L
Finally, we can calculate the concentration of OH- ions in the final solution:
Concentration of OH- ions = (moles of OH- ions) / (total volume of solution)
Concentration of OH- ions = 0.02475 moles / 0.3 L
Concentration of OH- ions ≈ 0.0825 M
Therefore, the concentration of the excess OH- ions in the solution is approximately 0.0825 M.