PRR asked in Science & MathematicsChemistry · 9 months ago

Titrations question?

A 25.0 cm3 sample of chlorinated water requires 27.2 cm3 of 0.104 mol dm-3 silver nitrate solution to be used to precipitate all of the Cl- in the sample. Determine the [Cl-] in the water sample.

What volume, in cm3, of 0.25 mol dm-3 sodium sulfate solution is required to precipitate all of the barium, as barium sulfate, from 12.5 cm3 of 0.15 mol dm-3 barium nitrate solution?

One test for lead (II) ions in water is to react them all with the chromate ion, forming lead(II) chromate, a bright yellow insoluble solid. How many grams of lead chromate can be formed from a 1.00 g sample of lead(II) nitrate added to 25.0 cm3 of 1.00 mol dm-3 potassium chromate solution?

1 Answer

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  • david
    Lv 7
    9 months ago

    A 25.0 cm3 sample of chlorinated water requires 27.2 cm3 of 0.104 mol dm-3 silver nitrate solution to be used to precipitate all of the Cl- in the sample. Determine the [Cl-] in the water sample.

    ===== 27.2 cm3 X 0.104 mol dm-3 X 1/25.0 cm3 = 0.113 mol dm-3 = [Cl-] <<<< answer

    What volume, in cm3, of 0.25 mol dm-3 sodium sulfate solution is required to precipitate all of the barium, as barium sulfate, from 12.5 cm3 of 0.15 mol dm-3 barium nitrate solution?

    ===== 12.5 cm3 X 0.15 mol dm-3 X 2 mol Na/1mole Ba X 1dm^3/0.25mol = 15 cm^3

    One test for lead (II) ions in water is to react them all with the chromate ion, forming lead(II) chromate, a bright yellow insoluble solid. How many grams of lead chromate can be formed from a 1.00 g sample of lead(II) nitrate added to 25.0 cm3 of 1.00 mol dm-3 potassium chromate solution?

    .... find molar mass of lead (II) chromate ... and MM of lead(II)nitrate

    [1.00g / MM lead(II)nitrate] X [1 / MM lead(II)chromate] = Answer 1 [moles lead(II)chromate]

    [0.0250 dm3 X 1.00 mol dm-3 potassium chromate] X [1 / MM lead(II)chromate] = Answer 2 [moles lead(II)chromate]

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    compare answer 1 and answer 2 ... the SMALLEST is the corret answer

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