Question

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.

Answer 1

concentration of CrO4²⁻ ions in the final solution = 6.53 × 10⁻⁵ mol /L

Step-by-step 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