Welp that sounds like a complicated prob...Good Luck!
Answer:
Explanation:
The solution contain 0.01 M concentration of Ba²⁺
0.01M concentration of Ca²⁺
Ksp ( solubility constant) for BaSO₄ = 1.07 × 10⁻¹⁰
Ksp for CaSO₄ = 7.10 × 10⁻⁵
(BaSO₄) = (Ba²⁺) (SO₄²⁻)
1.07 × 10⁻¹⁰ = 0.01 M (SO₄²⁻)
1.07 × 10⁻¹⁰ / 0.01 = ( SO₄²⁻)
1.07 × 10⁻⁸ M = ( SO₄²⁻)
so the minimum of concentration of concentration sulfate needed is 1.07 × 10⁻⁸ M
For CaSO₄
CaSO₄ = ( Ca²⁺) ( SO₄²⁻)
7.10 × 10⁻⁵ = 0.01 (SO₄²⁻)
(SO₄²⁻) = 7.10 × 10⁻⁵ / 0.01 = 7.10 × 10⁻³ M
so BaSO₄ will precipitate first since its cation (0.01 M Ba²⁺) required a less concentration of SO₄²⁻ (1.07 × 10⁻⁸ M ) compared to CaSO₄
b) The minimum concentration of SO₄²⁻ that will trigger the precipitation of the cation ( 0.01 M Ba²⁺) that precipitates first is 1.07 × 10⁻⁸ M
I believe it is testing your hypothesis and collecting data.
Incomplete question. The full question read;
You are analyzing water that is known to contain silver nitrate, AgNO3. You decide to determine the amount of silver nitrate using gravimetric analysis based on the reaction:
Ag+ + Cl– → AgCl
You add excess NaCl to a 100 ml sample of the water and find that 1.2 g of AgCl solid forms.
If you added excess NaCl to a 200 ml sample of water from the same source, how many grams of AgCl solid would you expect to form?
Answer:
<u>0.6 gram</u>
Explanation:
<em>Remember</em>, we were first told when NaCl is added to a 100 ml sample of the water it results in the formation of 1.2 g of AgCl.
Hence, if the volume of water is increased 2x to 200 ml from 100 ml, and NaCl is added to it, then the expected number of grams should be 0.6 (1.2g/2). That is, with increased volume, the amount of dissolution of AgCl is increased.
