0.4324 g sample of a potassium hydroxide-lithium hydroxide mixture requires 28.28 ml of 0.3520 M HCl for its titration to the equivalence point. What is the mass percent lithium hydroxide in this mixture?
12.34%
To find the mass percent of lithium hydroxide in the mixture, we need to calculate the moles of potassium hydroxide (KOH) and lithium hydroxide (LiOH), and then determine the mass percent of LiOH.
Here are the steps to solve the problem:
Step 1: Calculate the number of moles of HCl used in the titration.
Given:
- Volume of HCl used = 28.28 ml = 0.02828 L
- Molarity of HCl = 0.3520 M
Moles of HCl = molarity × volume
= 0.3520 M × 0.02828 L
Step 2: Calculate the moles of KOH used in the titration.
Since KOH and HCl react in a 1:1 ratio, the moles of KOH are the same as the moles of HCl used.
Moles of KOH = Moles of HCl
Step 3: Calculate the moles of LiOH in the mixture.
The moles of LiOH can be calculated by subtracting the moles of KOH used from the total moles of base in the mixture.
Total moles of base in the mixture = Moles of KOH + Moles of LiOH
Moles of LiOH = Total moles of base in the mixture - Moles of KOH
Step 4: Calculate the mass of LiOH.
Given:
- Mass of the sample = 0.4324 g
Mass of LiOH = Moles of LiOH × Molar mass of LiOH
Step 5: Calculate the mass percent of LiOH in the mixture.
Mass percent of LiOH = (Mass of LiOH / Mass of the mixture) × 100
Now, let's calculate each step:
Step 1: Moles of HCl used
Moles of HCl = 0.3520 M × 0.02828 L = 0.009955 moles HCl
Step 2: Moles of KOH used
Moles of KOH = 0.009955 moles HCl (since KOH and HCl react in a 1:1 ratio)
Step 3: Moles of LiOH in the mixture
Total moles of base in the mixture = Moles of KOH + Moles of LiOH
= 0.009955 moles KOH + Moles of LiOH
Moles of LiOH = Total moles of base in the mixture - Moles of KOH
Step 4: Mass of LiOH
Molar mass of LiOH = (atomic mass of Li) + (3 × atomic mass of O) + atomic mass of H
= (6.941 g/mol) + (3 × 16.00 g/mol) + (1.008 g/mol)
≈ 23.996 g/mol
Mass of LiOH = Moles of LiOH × Molar mass of LiOH
Step 5: Mass percent of LiOH in the mixture
Mass percent of LiOH = (Mass of LiOH / Mass of the mixture) × 100
Now, you can substitute the calculated values into the formula to find the mass percent of LiOH in the mixture.
To find the mass percent of lithium hydroxide in the mixture, we need to determine the amount of lithium hydroxide used in the titration.
Let's start by calculating the number of moles of HCl used in the titration:
Moles of HCl = Volume of HCl (in liters) * Concentration of HCl
Converting the volume of HCl from milliliters to liters:
Volume of HCl = 28.28 ml * (1 L / 1000 ml) = 0.02828 L
Substituting the values into the formula:
Moles of HCl = 0.02828 L * 0.3520 M = 0.009949 moles
Since potassium hydroxide (KOH) and lithium hydroxide (LiOH) react with HCl based on a 1:1 stoichiometric ratio, the number of moles of lithium hydroxide in the mixture is also 0.009949 moles.
Now, let's calculate the mass of lithium hydroxide:
Mass of LiOH = Moles of LiOH * Molar mass of LiOH
The molar mass of LiOH = Molar mass of Li + Molar mass of O + Molar mass of H
= (6.941 g/mol) + (15.999 g/mol) + (1.008 g/mol)
= 23.948 g/mol
Substituting the values into the formula:
Mass of LiOH = 0.009949 moles * 23.948 g/mol = 0.2377 grams
Finally, we can calculate the mass percent of lithium hydroxide:
Mass percent of LiOH = (Mass of LiOH / Mass of mixture) * 100
Since we are given that the sample weight is 0.4324 grams, substituting the values:
Mass percent of LiOH = (0.2377 g / 0.4324 g) * 100 = 55.03%
Therefore, the mass percent of lithium hydroxide in the mixture is approximately 55.03%.
You need two equations in two unknowns.
mass KOH + mass LiOH = 0.4324g
mols KOH + mols LiOH = 0.02828 x 0.3520
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Let X = mass KOH
and Y = mass LiOH
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X + Y = 0.4324
(X/molar mass KOH) + (Y/molar mass LiOH) = 0.02828 x 0.03520
Solve the two equations for X and Y, then,
mass % LiOH = (g LiOH/mass sample)*100 = ?
Post your work if you get stuck.