I am calculating the data for my lab and I am a bit puzzled on the following: 8)

Question

I am calculating the data for my lab and I am a bit puzzled on the following:

8) [OH-], equilibrium (mol/L)

9) [Ca2+}, equilibrium (mol/L)

10) molar solubility of Ca(OH)2 (mol/L)

11) average molar solubility of Ca(OH)2 (mol/L)

The questions prior to that, along with my anwers, were:

1) vol. Of Ca(OH)2 (mL) 25.0

2) conc of standerdized HCl solution (mol/L) 0.05

5) volume of HCl added (mL) 24.4

6) moles of HCl added 1.22x10^-3

7) moles of OH- in saturatued solutuin (mol) 4.88x10^-2

IF SOMEONE COULD PLEASE EXPLAIN HOW TO SOLVE 8-11) THAT WOULD BE GREAT!!!!!!

Explanation / Answer

You are titrating 25 ml of saturated Ca(OH)2 with 0.05M HCl? You used 1.14 x 10^-3moles HCl.

1 mole Ca(OH)2 reacts with 2 moles HCl so there were 5.7 x 10^-4 moles Ca(OH)2 in 25 ml or 5.7 x 10^-4 x 1000/25 = 0.0228 mol/litre (the molar solubility).

In the saturated solution

[Ca2+] = 5.7 x 10^-4 M and

[OH-] = 1.14x10^-3M

. Ksp = [Ca2+] x [OH-]^2.

=7.4 x 10^-10

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