The \"blank 1, 2, 3\" represent part a, b, c. Please show all work/steps thank y

Question

The "blank 1, 2, 3" represent part a, b, c.
Please show all work/steps thank you! A 1.217-g sample of commercial KOH contaminated by K2CO3 was dissolved in water, and the resulting solution was diluted to 500.0 mL. A 50.00-mL aliquot of this solution was treated with 40.00 mL of 0.05304 M HCI and boiled to remove CO2. The excess acid consumed 4.74 mL of 0.04983 M NaOH (phenolphthalein indicator). An excess of neutral BaCl2 was added to another 50.00-mL aliquot to precipitate the carbonate as BaCO3. The solution was then titrated with 28.56 mL of the acid to a phenolphthalein end point. Calculate the percentage KOH (Blank # 1), K2CO3 (Blank # 2), and H2O (Blank # 3) in the sample, assuming that these are the only compounds present.

Explanation / Answer

Given :

Mass of the sample in 500 mL of the solution = 1.127 g
So Mass of the sample in 50 mL of the solution will be 1.127 x (50/500) = 0.1127 g
Now let us consider the second titration first :

The excess BaCl2 will react with all the CO32- ions to give BaCO3 solid.

Therefore, the titration only involves the reaction between HCl and KOH.

HCl + KOH KCl + H2O
As per the stoichiometry of reaction 1 mole of HCl will react with 1 mole of KOH
No. of moles of HCl used = Molarity X Volume = 0.05304 x (28.56/1000) = 0.001515 moles
therefore No. of moles of KOH used = 0.001515 mol
we know that mMolar mass of KOH =56 g mol-1
Mass of KOH in 50 mL of solution = mol x (molar mass) = 0.001515 x 56 = 0.08484 g

3)
Let us consider the first titration now
Some of the HCl in 40 mL of the solution is used to react with KOH and K2­CO3 in 50 mL of the solution, and then the excess is titrated with NaOH solution.

HCl + KOH KOH + H2O
2HCl + K2CO3 2KCl + CO2 + H2O
Mole ratio HCl : K2CO3 = 2 : 1
HCl + NaOH NaCl + H2O
Mole ratio HCl : NaOH = 1 : 1

No. of moles of NaOH used for rection with HCl= Molarity X Volume = 0.04983 x (4.74/1000) = 0.0002362 mol
No. of moles of HCl used = 0.000236 mol
As calculated above, no. of moles of HCl used for reaction with KOH = 0.001515 mol
Total no. of moles of HCl used = MV = 0.05304 x (40/1000) = 0.002122 mol
No. of moles of HCl used for reaction with K2CO3 = 0.002122 - (0.001515 + 0.0002362) = 0.0003708 mol
So No. of moles of K2CO3 used = moles of HCl used / 2 = 0.0003708 x (1/2) = 0.0001854 mol
Molar mass of K2CO3 =138 g mol-1
Mass of K2CO3 in 50 mL of solution = mass x (molar mass) = 0.0001854 x 138 = 0.02559 g

Let us calculate the % now
In 50 mL of the solution:
Total mass of the sample = 0.1127 g
Mass of KOH = 0.08484 g
Mass of K2CO3 = 0.02559 g
% by mass of KOH = (0.08484/0.1127) x 100% = 75.28%
% by mass of K2CO3 = (0.02559/0.1127) x 100% = 22.71%
% by mass of H2O = 100% - (75.28% + 22.71%) = 2.01%

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