A 2.118 g sample of a solid mixture containing only potassium carbonate ( MM=138.2058 g/mol ) and potassium bicarbonate ( MM=100.1154 g/mol ) is dissolved in distilled water. A volume of 30.20 mL of a 0.763 M HCl standard solution is required to titrate the mixture to a bromocresol green end point. Calculate the weight percent of potassium carbonate and potassium bicarbonate in the mixture.
The HCl reacts completely with both as follows:
K2CO3 + 2HCl ==> 2KCl + H2O + CO2
KHCO3 + HCl ==> KCl + H2O + CO2
You set up two equations and solve them simultaneously.
Let x = grams K2CO3
and y = grams KHCO3.
eqn 1 is x + y = 2.118
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eqn 2 comes from
mols HCl to titrate K2CO3 + mols HCl to titrate KHCO3 = mols HCl used.
(2x/MM K2CO3) + (y/MM KHCO3) = mols HCl x L HCl
Solve these two equations simultaneously. When you get x and y then
% K2CO3 = (g K2CO3/wt sample)*100 = ?
% KHCO3 = (g KHCO3/wt sample)*100 = ?
Post your work if you get stuck.
@DrBob222
I am stuck,
x=1.590 g of k2CO3=> x+y=2.118
1.590+y=2.118 => y=0.528 g of KHCO3
then 0.528 g to moles
2x+0.00527 mole=0.0230 mol => x=1.225 g of KCO3
To calculate the weight percent of potassium carbonate and potassium bicarbonate in the mixture, we need to determine the moles of each compound and then convert them into weight percentages.
Let's break down the steps to calculate this:
Step 1: Calculate the moles of HCl used in the titration.
The molarity of the HCl solution is 0.763 M, and the volume used is 30.20 mL (which can be converted to liters).
moles of HCl = molarity × volume
moles of HCl = 0.763 M × 0.03020 L
moles of HCl = 0.0231 mol
Step 2: Calculate the moles of carbonates in the mixture.
Since the titration required HCl, which reacts with the carbonates, we can determine the moles of carbonates present.
From the balanced equation of the reaction:
HCl + K2CO3 → KCl + CO2 + H2O
we can see that 1 mole of HCl reacts with 1 mole of K2CO3.
So, moles of carbonates = moles of HCl used in the titration = 0.0231 mol.
Step 3: Calculate the moles of bicarbonates in the mixture.
We need to determine the excess moles of HCl used in the titration, which can be subtracted from the total moles of HCl used.
Total moles of HCl = moles of HCl used in the titration = 0.0231 mol.
Since 1 mole of HCl reacts with 1 mole of bicarbonates, the excess moles of HCl used in the titration will react with bicarbonates present.
moles of bicarbonates = total moles of HCl - moles of carbonates
moles of bicarbonates = 0.0231 mol - 0.0231 mol
moles of bicarbonates = 0 mol
Step 4: Calculate the moles of potassium carbonate and potassium bicarbonate in the mixture.
The molar masses of potassium carbonate and potassium bicarbonate are:
MM(K2CO3) = 138.2058 g/mol
MM(KHCO3) = 100.1154 g/mol
moles of potassium carbonate = moles of carbonates = 0.0231 mol
moles of potassium bicarbonate = moles of bicarbonates = 0 mol
Step 5: Calculate the weight percent of potassium carbonate and potassium bicarbonate in the mixture.
First, we need to find the weight of the mixture using its mass.
mass of mixture = 2.118 g
Now, we can calculate the weight percent of each compound using the following formulas:
weight percent of potassium carbonate = (moles of potassium carbonate × MM(K2CO3)) / mass of mixture × 100%
weight percent of potassium bicarbonate = (moles of potassium bicarbonate × MM(KHCO3)) / mass of mixture × 100%
In this case, the weight percent of potassium bicarbonate will be 0 since there were no bicarbonates present.
Let's calculate the weight percent of potassium carbonate:
weight percent of potassium carbonate = (0.0231 mol × 138.2058 g/mol) / 2.118 g × 100%
weight percent of potassium carbonate = 1.51%
So, the weight percent of potassium carbonate in the mixture is 1.51%.
To calculate the weight percent of potassium carbonate and potassium bicarbonate in the mixture, you need to determine the amount of each substance present in the sample.
First, calculate the number of moles of HCl used in the titration:
moles HCl = concentration of HCl (in M) × volume of HCl (in L)
= 0.763 M × 0.03020 L
= 0.02308 moles HCl
Since the balanced chemical equation between HCl and K2CO3 is 2 HCl + K2CO3 → 2 KCl + CO2 + H2O, we know that 2 moles of HCl react with 1 mole of K2CO3.
Therefore, the number of moles of K2CO3 in the sample is half of the moles of HCl used in the titration:
moles K2CO3 = 0.02308 moles HCl / 2
= 0.01154 moles K2CO3
Now, calculate the number of moles of HCl reacted with KHCO3, which is 1 mole of HCl for 1 mole of KHCO3.
The remaining moles of HCl are used to react with KHCO3:
moles KHCO3 = moles HCl - moles K2CO3
= 0.02308 moles HCl - 0.01154 moles K2CO3
= 0.01154 moles KHCO3
Next, calculate the weights of K2CO3 and KHCO3 in the sample:
weight K2CO3 = moles K2CO3 × molar mass K2CO3
= 0.01154 moles K2CO3 × 138.2058 g/mol
= 1.5959 g K2CO3
weight KHCO3 = moles KHCO3 × molar mass KHCO3
= 0.01154 moles KHCO3 × 100.1154 g/mol
= 1.151 g KHCO3
Now, calculate the weight percent of K2CO3 and KHCO3 in the mixture:
weight percent K2CO3 = (weight K2CO3 / total weight of sample) × 100
= (1.5959 g / 2.118 g) × 100
= 75.29%
weight percent KHCO3 = (weight KHCO3 / total weight of sample) × 100
= (1.151 g / 2.118 g) × 100
= 54.30%
Therefore, the weight percent of potassium carbonate in the mixture is approximately 75.29%, and the weight percent of potassium bicarbonate is approximately 54.30%.