A 100.0 mL solution containing aqueous HC1 and HBr was titrated with 0.1280 M Na
ID: 529828 • Letter: A
Question
A 100.0 mL solution containing aqueous HC1 and HBr was titrated with 0.1280 M NaOH. The volume of base required to neutralize the acid was 45.22 mL. Aqueous AgNO_3 was then added to precipitate the Cl^- and Br^- ions as AgCl and AgBr. The mass of the silver halides obtained was 0.9825 g. What is the molarity of the HBr in the original solution? Express your answer using four significant figures. What is the molarity of the HC1 in the original solution? Express your answer using four significant figures.Explanation / Answer
HBr and HCl are both monoprotic and react in the same ratio (1:1) with NaOH
HBr + NaOH ---> NaBr + H2O
HCl + NaOH ---> NaCl + H2O
So the moles of NaOH used will be the same as total moles of HBr + HCl in the mixture
moles NaOH = molarity x litres
= 0.1280 M x 0.04522 L
= 0.00578816 mol
Therefore moles HCl + moles HBr = 0.00578816 mol
There is one Cl- ion in each HCl and one Br- ion in HBr
Therefore total moles Br- + Cl- = total moles acid
= 0.00578816 mol
And, each AgBr and AgCl has only 1 Br- or Cl- so...
moles AgCl + moles AgBr = 0.00578816 mol
Now, we need to find what combination of moles of AgBr and AgCl = 0.9825 g
let moles AgBr = a
moles AgCl = total moles - moles AgBr = 0.00578816 - a
mass AgBr = molar mass x moles
= 187.77 g/mol x a mol
= 187.77a
mass AgCl = molar mass x moles
= 143.32 g/mol x (0.00578816 -a)
= 0.8295590912 - 143.32a
We know the total mass of Halide = 0.9825 g
and we know that mass AgBr + mass AgCl = 0.9825 g
so
187.77a + 0.8295590912 - 143.32a = 0.9825 g
solve for a
187.77a - 143.32a = 0.9825 - 0.8295590912
=>44.45 a = 0.1529409088
=>a = 0.1529409088 / 44.45
=>a= 0.00344074 mol
a = moles AgBr
moles AgBr = 0.00344074 mol
each AgBr has 1 Br-, so moles Br- (and thus HBr) = 0.00344074 mol
moles AgCl = total moles - moles AgBr
= 0.00578816 mol - 0.00344074 mol
= 0.00234742 mol
Each AgCl has 1 Cl- so moles Cl- (thus HCl) = 0.00234742 mol
Total volume was 100.0 mL = 0.1000 L
molarity = moles / Litres
Molarity HBr = 0.00344074 mol / 0.1000 L
= 0.0344074 M
Molarity HCl = 0.00234742 mol / 0.1000 L
= 0.0234742 M
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