Question

In: Chemistry

A 21.466 g sample of aqueous waste leaving a fertilizer manufacturer contains ammonia. The sample is...

A 21.466 g sample of aqueous waste leaving a fertilizer manufacturer contains ammonia. The sample is diluted with 70.668 g of water. A 11.619 g aliquot of this solution is then titrated with 0.1062 M HCl. It required 30.75 mL of the HCl solution to reach the methyl red endpoint. Calculate the weight percent NH3 in the aqueous waste.

Solutions

Expert Solution

We know that

No. mole of = molarity Volume of solution in liter

no mole of HCl = 0.1062M 0.03075L = 0.00326565 mole

reaction take place between HCl and NH3 as follow

HCl(aq) + NH3(aq)  NH4Cl(aq)

1:1 HCl and NH3 required then for 0.00326565 mole of HCl required 0.00326565 mole of NH3

That mean 11.619 gm of aliquot contain 0.00326565 mole of NH3

Total mass of NH3 solution = 21.466gm +70.668 = 92.134 gm

11.619 gm of aliquot contain 0.00326565 mole of NH3 then 92.134 gm contain

92.1340.00326565 / 11.619 = 0.02589 mole of NH3

molar mass of NH3 = 17.031 gm/mol that mean 1 mole of NH3 = 17.031 gm then 0.02589 mole of NH3 =

0.0258917.031 = 0.4409 gm of NH3

92.134 gm of solution contain 0.4409 gm Ammonia

weight % of solute = mass of solute100 / mass of solution

weight % of NH3 = 0.4409100/92.134 = 0.4785%


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