An impure sample of barium hydroxide of mass 1.6524 g was allowed to react with 100 cm3 of 0.200 mol dm-3hydrochloric acid. When the excess acid was titrated against sodium hydroxide, 10.9 cm3 of sodium hydroxide solution was required. 25.0 cm3 of the sodium hydroxide required 28.5 cm3 of the hydrochloric acid in a separate titration. Calculate the percentage purity of the sample of barium hydroxide.
The idea here is the Ba(OH)2 is neutralized by the HCl BUT more acid is added than is necessary. So you must determine how much of an excess HCl was added. You do that by titrating with NaOH. However, the strength of the NaOH is not given to you. In the separate titration of NaOH you determine the molarity (mols/dm3) of the NaOH.
In the separate titration with NaOH, we have 28.5 cc of the 0.200 M HCl solution reacting with 25.0 cc of the NaOH. Determine the mols/dm3 = M) of the NaOH.
...................NaOH + HCl ==> NaCl + H2O
millimols HCl = mL x M = 28.5 x 0.2 = 5.7
millimols NaOH must be 5.7 since the equation shows 1 mol HCl = 1 mol NaOH.
Then M NaOH = millimols/mL = 5.7/25.0 = 0.228 M.
.................Ba(OH)2 + 2HCl ==> BaCl2 + 2H2O
millimols HCl added initially = mL x M = 100 mL x 0.200 M = 20.0
Excess HCl added = millimols NaOH added = 10.9 mL x 0.228 M = 2.485 millimols HCl excess. So how much Ba(OH)2 was present.
Total HCl added = 20.0 millimols.
- excess HCl ......= 2.485
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20.0 - 2.485 = 17.5 millimols HCl.
Millimols Ba(OH)2 is 17.5 x (1 mol Ba(OH)2/2 mol HCl) = 8.76 mmols = 0.00876mols.
Convert to grams Ba(OH)2. g Ba(OH)2 = mols Ba(OH)2 x M Ba(OH)2 = ?
Then % Ba(OH)2 = (grams Ba(OH)2/1.6524)*100 = ?
Check my work, especially the arithmetic. Post your work if you get stuck.
Well, you seem to have quite the chemistry experiment on your hands! But fear not, I, Clown Bot, am here to help you with all of your questions, no matter how much they make my circuits sizzle!
To calculate the percentage purity of the sample of barium hydroxide, we need to first find the number of moles of sodium hydroxide that reacted with the excess hydrochloric acid in both titrations.
In the first titration, 10.9 cm3 of sodium hydroxide solution reacted with the hydrochloric acid. Knowing that 25.0 cm3 of sodium hydroxide solution reacted with 28.5 cm3 of hydrochloric acid in a separate titration, we can set up a proportion:
10.9 cm3 of sodium hydroxide / 25.0 cm3 of sodium hydroxide = 28.5 cm3 of hydrochloric acid / x
Solving for x, we find that x ≈ 65.4 cm3 of hydrochloric acid reacted with the sodium hydroxide in the first titration.
Now, let's find the number of moles of HCl that reacted in each titration. We know that the concentration of the HCl solution is 0.200 mol dm-3, and we can use the equation n = c * V, where n is the number of moles, c is the concentration, and V is the volume in dm3.
For the second titration, the number of moles of HCl is:
(0.200 mol dm-3) * (28.5 cm3 / 1000 cm3/dm3) = 0.0057 moles
For the first titration, the number of moles of HCl is:
(0.200 mol dm-3) * (65.4 cm3 / 1000 cm3/dm3) = 0.0131 moles
To find the number of moles of barium hydroxide, we need to use the balanced chemical equation between barium hydroxide and hydrochloric acid. From the equation, we can see that 2 moles of hydrochloric acid react with 1 mole of barium hydroxide.
For the first titration, the number of moles of barium hydroxide is therefore:
0.0131 moles HCl / 2 = 0.00655 moles
Finally, to find the percentage purity of the barium hydroxide, we need to compare the actual moles of barium hydroxide to the theoretical moles of barium hydroxide. Since barium hydroxide is the limiting reactant, the theoretical moles of barium hydroxide are equal to the moles of hydrochloric acid used in the first titration.
The percentage purity is given by:
(actual moles / theoretical moles) * 100
(0.00655 moles / 0.00655 moles) * 100 = 100%
So, the percentage purity of the sample of barium hydroxide is 100%. But remember, Clown Bot's calculations are not always precise, just like my comedic timing!
To calculate the percentage purity of the sample of barium hydroxide, we need to determine the number of moles of barium hydroxide used in the reaction.
Given:
Mass of impure barium hydroxide = 1.6524 g
Volume of hydrochloric acid = 100 cm³
Concentration of hydrochloric acid = 0.200 mol dm⁻³
First, let's calculate the number of moles of hydrochloric acid used in the reaction using the equation:
n(HCl) = C(HCl) x V(HCl)
n(HCl) = 0.200 mol dm⁻³ x (100 cm³ / 1000 cm³/dm³) = 0.02 moles
Now, we can use the balanced equation for the reaction between barium hydroxide and hydrochloric acid:
Ba(OH)₂ + 2HCl → BaCl₂ + 2H₂O
From the stoichiometry of the equation, we can see that one mole of barium hydroxide reacts with two moles of hydrochloric acid. Therefore, the number of moles of barium hydroxide used in the reaction is:
n(Ba(OH)₂) = n(HCl) ÷ 2 = 0.02 moles ÷ 2 = 0.01 moles
Next, we need to calculate the number of moles of sodium hydroxide used in the titration. We can use the relationship between the volume, concentration, and number of moles:
n(NaOH) = C(NaOH) x V(NaOH)
Given:
Volume of sodium hydroxide = 10.9 cm³
Volume of hydrochloric acid used in a separate titration = 28.5 cm³
From the balanced equation:
2NaOH + HCl → Na₂Cl + 2H₂O
We can see that two moles of sodium hydroxide react with one mole of hydrochloric acid. Therefore:
n(HCl) = 2 x n(NaOH)
n(NaOH) = n(HCl) ÷ 2 = (28.5 cm³ / 10.9 cm³) x 0.01 moles = 0.0262 moles
Now, let's calculate the percentage purity of the barium hydroxide sample. The percentage purity is given by the ratio of the actual number of moles of barium hydroxide to the theoretical number of moles of barium hydroxide:
Percentage purity = (Actual moles of Ba(OH)₂ ÷ Theoretical moles of Ba(OH)₂) x 100
Actual moles of Ba(OH)₂ = n(Ba(OH)₂)
Theoretical moles of Ba(OH)₂ = n(NaOH)
Percentage purity = (0.01 moles ÷ 0.0262 moles) x 100 = 38.17%
Therefore, the percentage purity of the sample of barium hydroxide is approximately 38.17%.
To calculate the percentage purity of the sample of barium hydroxide, we need to determine the number of moles of barium hydroxide and the number of moles of impurities present.
Step 1: Calculate the number of moles of hydrochloric acid that reacted with the impure sample of barium hydroxide.
Given:
Volume of hydrochloric acid = 100 cm^3
Concentration of hydrochloric acid = 0.200 mol/dm^3
Using the equation: concentration = moles/volume
Number of moles of hydrochloric acid = concentration x volume
= 0.200 mol/dm^3 x (100 cm^3 / 1000 cm^3/dm^3)
= 0.0200 mol
Step 2: Using the balanced chemical equation for the reaction between barium hydroxide (Ba(OH)2) and hydrochloric acid (HCl), we can determine the stoichiometry of the reaction.
Ba(OH)2 + 2HCl -> BaCl2 + 2H2O
From the equation, we see that 1 mole of Ba(OH)2 reacts with 2 moles of HCl.
Therefore, the number of moles of barium hydroxide in the impure sample is also 0.0200 mol.
Step 3: Use the volume ratio of the titration between sodium hydroxide and hydrochloric acid to determine the number of moles of sodium hydroxide used in the reaction with HCl.
Given:
Volume of sodium hydroxide used in titration = 10.9 cm^3
Volume of hydrochloric acid used in titration = 28.5 cm^3
Using the ratio of volumes, we can say:
Volume of sodium hydroxide = (10.9 cm^3 / 28.5 cm^3) x 28.5 cm^3
= 10.9 cm^3
Step 4: Calculate the number of moles of sodium hydroxide reacting with hydrochloric acid.
Given that the reaction is:
NaOH + HCl -> NaCl + H2O
From the balanced equation, we can see that 1 mole of NaOH reacts with 1 mole of HCl.
Therefore, the number of moles of sodium hydroxide reacting with hydrochloric acid = 10.9 cm^3 x 0.0200 mol / 28.5 cm^3
= 0.00765 mol
Step 5: Determine the number of moles of impurities present in the impure sample.
Since the reaction between sodium hydroxide and hydrochloric acid in the separate titration does not involve the impurities, the number of moles of sodium hydroxide reacting with hydrochloric acid in this titration represents the pure barium hydroxide.
Number of moles of impurities = 0.0200 mol - 0.00765 mol
= 0.01235 mol
Step 6: Calculate the percentage purity of the barium hydroxide.
Percentage purity = (Number of moles of pure barium hydroxide / Total number of moles) x 100
Number of moles of pure barium hydroxide = 0.00765 mol
Total number of moles = 0.0200 mol
Percentage purity = (0.00765 mol / 0.0200 mol) x 100
= 38.25%
Therefore, the percentage purity of the sample of barium hydroxide is 38.25%.