I am confused about where I am going wrong here...
Suppose that 1.331 g of impure barium hydroxide is dissolved in enough water to produce 250. mL of solution and that 35.0 mL of this solution is titrated to the stoichiometric point with 17.6 mL of 0.0935 M HCl(aq). What percentage of the impure sample was barium hydroxide?
I found the concentration of the 35 mL Ba(OH)2 soln to be .047M and I know that there is a 1:2 ratio for Ba(OH)2 and HCl. I attempted to find the amount of moles of HCl (which I thought was .0016) and dividing that by 2 to get .000823 mol Ba(OH)2. I then multiplied that by MW 171.34 g/mol to get .141 g. I then divided this by the total given (1.331g) and found my percentage to be 10.59, which is incorrect. I've tried doing this a few times, playing around with the values and methods but can't get it right. As always, help is much appreciated!
I think the error is in the first step. Since the sample being analyzed is not pure Ba(OH)2, the molarity of an impure sample doesn't mean anything. Here is the way I would approach it; all the steps you already know.
1. I like to work in millimiles so let me do that and if you don't like it just divide mL by 1000 and mmoles by 1000. (mL x M = millimoles; M = mmoles/mL.
mmoles HCl = 17.6 mL x 0.0935 = about 1.6
mmoles Ba(OH)2 = 1.6 x 1/2 = about 0.8 because the equation is
2HCl + Ba(OH)2 ==> etc
grams Ba(OH)2 = moles x molar mass = about 0.0008 x 171.34 = about 0.14g
percent = (mass Ba(OH)2/mass sample)*100
mass sample = 1.331 x (35 mL/250 mL) = about 0.19 grams.
%Ba(OH)2 = (about 0.14 g Ba(OH)2/about 0.19)*100 = about 75% Ba(OH)2.
Check my thinking. Check my arithmetic.
[Note: just for your information, I do all of this in one step.
(mL x N x milliequivalent weight/mass sample *100 = % but the IUPAC, in all of it's wisdom, has declared equivalent weight obsolete and it isn't taught anymore. Gravimetric factors have been declared obsolete, also, and aren't taught anymore. I see some mention (a sentence or two) in some texts about normality but nothing about gravimetric factors.
To find the percentage of the impure sample that is barium hydroxide, you can use the following steps:
Step 1: Calculate the number of moles of HCl used in the titration.
To do this, you can use the formula: moles = concentration × volume.
Given concentration: 0.0935 M
Given volume: 17.6 mL = 0.0176 L
Moles of HCl = 0.0935 M × 0.0176 L = 0.001646 mol
Step 2: Since the stoichiometric ratio between Ba(OH)2 and HCl is 1:2, you need to multiply the moles of HCl by 2 to get the moles of Ba(OH)2 reacted.
Moles of Ba(OH)2 = 0.001646 mol × 2 = 0.003292 mol
Step 3: Determine the molar mass of Ba(OH)2.
Barium (Ba) has a molar mass of approximately 137.33 g/mol.
Hydrogen (H) has a molar mass of approximately 1.01 g/mol.
Oxygen (O) has a molar mass of approximately 16.00 g/mol.
Molar mass of Ba(OH)2 = (137.33 g/mol) + 2 × [(1.01 g/mol) + (16.00 g/mol)] = 137.33 g/mol + 34.02 g/mol = 171.35 g/mol
Step 4: Calculate the mass of Ba(OH)2 reacted.
Mass of Ba(OH)2 = moles of Ba(OH)2 × molar mass of Ba(OH)2
Mass of Ba(OH)2 = 0.003292 mol × 171.35 g/mol = 0.564 g
Step 5: Calculate the percentage of the impure sample that is barium hydroxide.
Percentage = (mass of Ba(OH)2 / total mass of impure sample) × 100%
Percentage = (0.564 g / 1.331 g) × 100% = 42.40%
Therefore, the percentage of the impure sample that is barium hydroxide is approximately 42.40%.
To determine the percentage of impure sample that is barium hydroxide, we need to first find the moles of barium hydroxide in the 35.0 mL solution that reacted with the HCl.
1. Start by finding the moles of HCl that reacted with the barium hydroxide:
Moles of HCl = (volume of HCl solution in L) x (molarity of HCl solution)
Convert the volume from mL to L:
Volume of HCl solution = 17.6 mL = 0.0176 L
Substitute the values:
Moles of HCl = 0.0176 L x 0.0935 mol/L = 0.001644 mol HCl
2. Since the stoichiometric ratio between barium hydroxide (Ba(OH)2) and HCl is 1:2, the moles of barium hydroxide is half the moles of HCl:
Moles of Ba(OH)2 = (1/2) x Moles of HCl = 0.000822 mol Ba(OH)2
3. Now, calculate the mass of barium hydroxide in the 35.0 mL solution:
Mass of Ba(OH)2 = Moles of Ba(OH)2 x Molar mass of Ba(OH)2
Molar mass of Ba(OH)2 = (1 x Molar mass of Ba) + (2 x Molar mass of O) + (2 x Molar mass of H)
Molar mass of Ba = 137.33 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of H = 1.01 g/mol
Substitute the values:
Molar mass of Ba(OH)2 = (1 x 137.33) + (2 x 16.00) + (2 x 1.01) = 171.34 g/mol
Now, calculate the mass:
Mass of Ba(OH)2 = 0.000822 mol x 171.34 g/mol = 0.141 g Ba(OH)2
4. Finally, calculate the percentage:
Percentage of barium hydroxide = (Mass of Ba(OH)2 / Mass of impure sample) x 100
Mass of impure sample = 1.331 g
Substitute the values:
Percentage of barium hydroxide = (0.141 g / 1.331 g) x 100 = 10.59%
So, the percentage of the impure sample that is barium hydroxide is approximately 10.59%.