7) b. Based on the comparison of size and Zeff, which of the three values of Ksp

Question

7) b. Based on the comparison of size and Zeff, which of the three values of Ksp listed in question 5 would Coulomb’s Law predict for beryllium hydroxide’s Ksp? c. Based on the comparison of size and Zeff, which of the three values of Ksp listed in question 6 would Coulomb’s Law predict for barium hydroxide’s Ksp? Question 5:Based on your results, which of the following values of Ksp do you think is most likely for beryllium hydroxide, Be(OH)2? Explain your choice. • 3.0 × 108 • 7.0 x 1022 • 2.0 x 103 Question 6: Based on your results, which of the following values of Ksp do you think is most likely for barium hydroxide, Ba(OH)2? Explain your choice. • 3.0 × 108 • 7.0 x 1022 • 2.0 x 103

Explanation / Answer

Ksp represents the solubility of a compound

It is derived as follows
for example,

NaCl (s) --> Na+ (aq) + Cl- (aq)

Ksp = [P] / [R]
Ksp = [Na+][Cl-] / 1

so, if Ksp is big, then there is more products at equilibrium (thus, the compound is MORE SOLUBLE).
If Ksp is small, then there is more reactants at equilibrium (thus, the compound is LESS soluble)

Judging from your solubility rules dealing with hydroxide ions

"6. Most hydroxide salts are only slightly soluble. Hydroxide salts of Group I elements are soluble. Hydroxide salts of Group II elements (Ca, Sr, and Ba) are slightly soluble. Hydroxide salts of transition metals and Al3+ are insoluble. Thus, Fe(OH)3, Al(OH)3, Co(OH)2 are not soluble"


-- we know from this that Be(OH)2 is not soluble, so Ksp is going to be as small as possible
-- we know from this that Ba(OH)2 IS soluble marginally, so Ksp is going to be bigger (but not that big!)

From this;

(7b and 5) 7.0 x 10^-22
(7c and 6) 2.0 x 10^-3

Zeff is defined as the effective nuclear charge.

Zeff = Z - S
Where Z is the number of protons in the nucleus and S is essentially the number of nonvalence electrons

Coulomb's law defines the electrostatic force of interaction between two point charges is directly proportional to the product of both charges and inversely proportional to the square of the distance between them;

F = k Qq /(r^2)

We can relate this equation to our question; Zeff is related to the charges in the equation, Qq; R is related to the size of atom due to it being distance-related

Beryllium has a Zeff of 1.5 (Q's) and is considerably smaller than Barium (r is small)

Barium has a Zeff of 2 (Q's) and is considerably larger than Beryllium (r is big)


7 b and 7c should just verify with physics that the solubility rules are correct.

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