8. 10 g of a compound with logP of 0.48 is dissolved in 150 mL of water. To the

Question

8. 10 g of a compound with logP of 0.48 is dissolved in 150 mL of water. To the nearest gram, how much compound is extracted into 200 mL of octan-1-ol from the aqueous solution when mixed in a separatory funnel? a. 2 b. 4 c. 6 d. 8 e. not a.-d. 9. To the nearest tenth, what is the pH of a 3.0 M solution of compound with pk, of 1.01? a. 1.0 b. 0.8 C. 0.6 8.0.4 e. not a.-d. 10. The equilibrium constant for the reaction in question 11 is given by? a. 10pka(forward)+pka (reverse) b. 10pka(forward)-pka(reverse) c. 10-pka(forward)+pka(reverse) 10-pka(forward)-pka(reverse) e, not a.-d. 11. Given the following pka's and initial concentration of reactants, to the nearest tenth, how much water is produced in the following reaction? pk, 16.6 pk, 15.7 Li**OH - CH + 4,0 CH + H20 3.0 M 3.0 M

Explanation / Answer

Ans 9

assume a monoprotic weak acid

Dissociation reaction with ICE TABLE

HA(aq)+H2O(l)A(aq)+H3O+(aq)

I 3

C - x    +x +x

E 3-x    x x

Equilibrium constant expression of the reaction

Ka=[A][H3O+]/[HA]

Now, given that

pKa = 1.01

pKa = -log Ka

Ka = 10^-1.01 = 0.0977

0.0977 = x2 / (3-x)

0.2931 - 0.0977 x - x2 = 0

x = 0.4947

[H3O+] = 0.4947

pH = - log[H3O+] = - log (0.4947)

pH = 0.3

Option e is the correct answer

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