For part B I worked with x as a variable and got that x= 0.0111199. Is that it?
ID: 965939 • Letter: F
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For part B I worked with x as a variable and got that x= 0.0111199. Is that it? Or should I say 4x^3 ions? The accepted value of the Ks for calcium hydroxide was given in this lab. Compare it with Comrment on the error and results.k6P- · ible sources of experimental error 5. The illustrations in this lab of saturated solutions of NaOH and Ca(OH)2 are given earlier. a 1625 How can you tell from all three representations that these are saturated solutions? You are given this note: "The sol a. 6 b, Y ou are given this note: "The solubility difference between NaOH and Ca(OH)2 is actually much greater. The solubility of Ca(OH)2 in particular is shown much igher here to include a few ions in solution. The solubility is actually much, much less." How many ions are in 15.0 mL of a saturated calcium hydroxide solution? hi How many ions are in 15.0 mL of a saturated sodium hydroxide solution? density.) hydroxide? hydroxide? How should the illustrations be altered to reflect this? c. d. How many water molecules are in 15.0 mL of water? (Use 1.00 g mL-1 as the e. What is the ratio of ions to water molecules for a saturated solution of sodium f. What is the ratio of ions to water molecules for a saturated solution of calcium g. The Ksp of Mg(OH)2 is 1.8 x 10-l1. Based on their Ksp values, show how magnesium hydroxide and calcium hydroxide are different on the particle level. What is the species in greatest concentration in both of these solutions? 6. RpB.I ARORATORY ASSIGNMENTExplanation / Answer
(b) Ca(OH)2 ---> Ca+2 + 2 OH-
Ksp = 5.5*10^-6 = [Ca+2] * [OH-]2 = x * (2x)2
5.5*10^-6 = 4 * x3
x = 0.0111199 M
So, there are:
moles of Ca+ = 0.0111199 M * 0.015 L = 0.000167 moles Ca+
moles of OH- = (2*0.0111199 M ) * 0.015 L = 0.000333 moles OH-
total moles = 0.000167 + 0.000333 = 0.0005 moles.
Remember Avogadro's Number: 6.02 * 10^23 representative particles. For one mole of substance, there are 6.02 * 10^23 representative particles. These particles can be atoms, molecules, ions, etc.
total ions = 0.0005 moles * 6.02 * 10^23 ions / 1 mol = 0.00301*10^23 ions
(c) NaOH ----> Na+ + OH-
Ksp = 110 = [Na+] * [OH-] = x2
x = 10.49 M
Moles of Na = Moles of OH-
Total moles = 2 * M * Volume = 2 * 10.49 * 0.015 L = 0.3147 moles
ions = 0.3147 moles * 6.02 * 10^23 ions / 1 mol = 1.89*10^23 ions
(d) density = mass /volume
mass = density * volume = 1 g/ml * 15 ml = 15 g
Molar mass of water = 18 g/mol
moles of water = mass/ molar mass = 15 g / 18 g/mol = 0.83 moles
0.83 moles H2O * 6.02 * 10^23 molecules H2O / 1 mol H2O
= 5* 10^23 molecules
(e) 1.89*10^23 ions / 5* 10^23 molecules = 0.378
(f) 0.00301*10^23 ions / 5* 10^23 molecules = 0.000602
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