Question

Suppose an EPA chemist tests a 250.mL sample of groundwater known to be contaminated with nickel(II) chloride, which would react with silver nitrate solution like this: NiCl2(aq) + 2AgNO3(aq) → 2AgCl(s) + NiNO32(aq) The chemist adds 15.0mM silver nitrate solution to the sample until silver chloride stops forming. He then washes, dries, and weighs the precipitate. He finds he has collected 5.8mg of silver chloride. Calculate the concentration of nickel(II) chloride contaminant in the original groundwater sample. Round your answer to 2 significant digits.

Answer 1

0.0010 w/v %

Step-by-step explanation:

Based on the reaction:

NiCl₂(aq) + 2AgNO₃(aq) → 2AgCl(s) + Ni(NO₃)₂(aq)

Finding the moles of AgCl produced we can find the moles of NiCl in the reaction medium and its mass:

Moles AgCl -Molar mass: 143.32g/mol-:

5.8mg = 5.8x10⁻³g * (1mol / 143.32g) = 4.047x10⁻⁵ moles AgCl

Moles NiCl₂:

4.047x10⁻⁵ moles AgCl * (1mol NiCl₂ / 2mol AgCl) = 2.023x10⁻⁵ moles NiCl₂

Mass NiCl₂ -Molar mass: 129.60g/mol-:

2.023x10⁻⁵ moles NiCl₂ * (129.60g / mol) = 2.62x10⁻³g of NiCl₂ are produced.

And the concentration in w/v% is:

2.62x10⁻³g NiCl₂ / 250mL * 100 =

0.0010 w/v %