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Answers and explanations for calculating equilibrium constants and ph values for given acids and bases using provided data. Topics include the relationship between ka and k, the calculation of poh, and the use of ice tables for weak bases.
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A) For nitruos acid, HNO 2 , Ka = 4.5 10-4^. What is the equilibrium constant for the following reaction?
NO 2 -^ + H 3 O +^ ⇔ HNO 2 + H 2 O
Answer: We have Ka , so we write the equation for Ka.
HNO 2 + H 2 O ⇔ NO 2 -^ + H 3 O +
This is just the inverse of the equation given in the problem, therefore K=1/Ka K= 2.2 10 3
Be careful that the given reaction is NOT the reaction of Kb for the conjugate base NO 2 - , Kb reaction: NO 2 -^ + H 2 O ⇔ HNO 2 + HO-
and you CANNOT use here K (^) w=Ka .Kb in this case.
B) A 0.001 aqueous solution of HClO 4 (Ka = large) has a pOH of _______.
Answer: It is a strong acid, therefore [acid] = [H 3 O +^ ] = 0.
pH = -log [H 3 O +^ ] = 3
We know that pH + pOH = 14,
pOH = 14 – pH = 14 – 3 = 11
C) Carbonate ion, CO 3 2-^ is a base, with Kb = 4.3 10-4^. Calculate the approximate [OH-] (not using the quadratic formula) for a 0.25 M solution of CO 3 Na 2.
Answer: carbonate is a weak base (Kb < 1). Therefore the [OH-] < [base]. We need to use an ICE table.
CO 3 2-^ + H 2 0 ⇔ HCO 3 -^ + HO - Initial (M) 0.25 -- -- Change (M) -x x x Equilibrium (M) 0.25 – x x x
Kb = [HCO3-] [HO - ] / [CO 3 2-] = x.x / (0.25 –x)
Since the initial concentration of carbonate is higher than 100. Kb , we can use the approximation that [CO 3 2-] 0 – [OH-] ≅ [CO 3 2-] 0 , where [CO 3 2-] 0 =0.25 M and [OH-]= x,
then
Kb = x^2 / 0.25; note: √(x) = x1/
x = √(Kb .0.25) = √(4.3 10-4^ .0.25) = √(0.000107) = 0.