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23 October, 03:08

39.20 mL of 0.5000 M AgNO3 is added to 270.00 mL

ofwater which contains 5.832 g K2CrO4. A

redprecipitate of Ag2CrO4 forms. What is

theconcentration, in mol/L, of

unprecipitatedCrO42-? Be sure to enter

the correct numberof significant figures. Assume

Ag2CrO4is completely insoluble.

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Answers (1)
  1. 23 October, 03:19
    0
    concentration of CrO4²⁻ ions in the final solution = 6.53 * 10⁻⁵ mol / L

    Explanation:

    First we calculate the number of moles of AgNO₃:

    number of moles = molar concentration * volume

    number of moles = 0.5 * 39.20 = 19.6 mmoles = 0,0196 moles AgNO₃

    Then we calculate the number of moles of K₂CrO₄:

    number of moles = mass / molar weight

    number of moles = 5.832 / 194 = 0.03 moles K₂CrO₄

    The chemical reaction will look like this:

    2 AgNO₃ + K₂CrO₄ → Ag₂CrO₄ + 2 KNO₃

    Now we devise the following reasoning:

    if 2 moles of AgNO₃ are reacting with 1 mole of K₂CrO₄

    then 0,0196 moles of AgNO₃ are reacting with X moles of K₂CrO₄

    X = (0.0196 * 1) / 2 = 0.0098 moles of K₂CrO₄

    now the the we calculate the amount of unreacted K₂CrO₄:

    unreacted K₂CrO₄ = 0.03 - 0.0098 = 0.0202 moles

    now the molar concentration of CrO4²⁻ ions:

    molar concentration = number of moles / solution volume (L)

    molar concentration = 0.0202 / (39.20 + 270) = 6.53 * 10⁻⁵ mol / L
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