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# 2015 AP Chemistry free response 4

Calculating the molar solubility of calcium hydroxide in a 0.10 M calcium nitrate solution. From 2015 AP Chemistry free response 4.

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• Ca(NO₃)₂ is already in the solution. So how come we are not including the concentration of NO₃⁻ in the equation for Ksp?
• It is because Ksp only specifies the concentration relations between Ca2+ and (OH)- regardless of other substances.
• in writing the balanced equation I ended up with:
Ca(OH)2 + H2O ---- CaO + 2H20
why is it wrong?
• Because they specifically asked for the dissolution reaction. When ionic compounds dissolve in water they turn into the ions that make it up, which in this case is Ca^2+ ions and OH^- ions.
• In part (a), why is it a double arrow? I thought that Ca(OH)2 is a strong base and would dissociate completely .
• Regardless of it being a strong base, Ksp expressions are written using double headed arrows
• what is molar solubility in b part
• around , how did you already have 0.10M of Ca+2 ions?
• You are starting with 0.10 mol/L Ca(NO₃)₂.
This is a strong electrolyte. It dissociates completely in solution.
Thus, 0.10 mol Ca(NO₃)₂ forms 0.10 mol Ca²⁺ and 0.20 mol NO₃⁻.
Then you add the Ca(OH)₂, which produces more calcium ions.
• So Sal essentially neglects the x in (x+.1), but how did he know that the calcium from the calcium hydroxide would be significantly less than .1M, what if someone had put in 5 times as much Ca(OH)2 as they had Ca(NO3)2, where did he get the intuition that x<<.1M?
• A rule of thumb is that if x is less than 5% of the number from which it is added or subtracted then the "small x" approximation is valid. Otherwise, it's necessary to solve the quadratic equation. In this video, x was 1.8% of 0.1 so the approximation was valid.
(1 vote)
• I wanted to calculate the molar solubility of CaCO3 in .050 M CaCl2 (Ksp for CaCO3 is 3.8 x 10^-9). I followed the steps in this video but with using the different numbers/exponents and got 2.8 x 10^-4, but the answer choices it gave were: a) 7.6 x 10^-8 b) 1.9 x 10^-7 c) 6.2 x 10^-5 d) 8.3 x 10^-5 e) 5.0 x 10^-2. What did I do wrong?
• I can't tell where you went wrong. You may simply have made a calculation error.
The correct answer is c) 6.2 × 10⁻⁵ mol/L.
CaCO₃ ⇌ Ca²⁺ + CO₃²⁻; Ksp = 3.8 × 10⁻⁹
_________x__x+0.050
Ksp = [Ca²⁺][CO₃²⁻] = 3.8 × 10⁻⁹
x(x+0.050) = 3.8×10⁻⁹
0.050/3.8×10⁻⁹ = 1.4 ×10⁷ >> 400; ∴ x ≪ 0.050
x² = 3.8×10⁻⁹
x = 6.2× 10⁻⁵
[CaCO₃] = [Ca²⁺] = x mol/L = 6.2 × 10⁻⁵ mol/L
(1 vote)
• at , why does Khan say "but that's going to be constant
because if there's any of the calcium hydroxide that gets dissolved, well the volume goes down,and if any gets formed, the volume goes up. So the concentration stays constant."
I don't think the logic holds up here: clearly the concentration is fluctuating, but then Khan says it stays constant.
(1 vote)
• The reactions where calcium hydroxide is dissolving and solidifying are happening simultaneously. So any amount of solid calcium hydroxide which dissociates into ions is immediately replaced by ions in solution solidifying back onto the block of sodium hydroxide. Therefore the concentration of the solid calcium hydroxide remains constant because the two rates are equal.

Hope that helps.
(1 vote)
• On part C (around ) is it necessary to label the H's and O's or the + and - sides of H2O, or can you get full credit without doing that?
(1 vote)
• So why did we have to include the 0.1M Ca(NO3)2 in our calculation in b? Does the fact that we already have some Ca in the solution decrease the solubility of our Ca(OH)2 ?
(1 vote)