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Equilibrium Constants and pH Calculations for Acids and Bases, Study Guides, Projects, Research of Chemistry

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.

Typology: Study Guides, Projects, Research

2021/2022

Uploaded on 09/12/2022

stefan18
stefan18 🇺🇸

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A) For nitruos acid, HNO2, Ka = 4.5 10-4. What is the equilibrium constant for the
following reaction?
NO2- + H3O+ HNO2 + H2O
1) 2.2 10-11
2) 2.2 10-3
3) 2.2 10 3
4) 4.5 10-10
5) None of the above
Answer:
We have Ka, so we write the equation for Ka.
HNO2 + H2O NO2- + H3O+
This is just the inverse of the equation given in the problem, therefore K=1/Ka
K= 2.2 103
Be careful that the given reaction is NOT the reaction of Kb for the conjugate
base NO2-,
Kb reaction: NO2- + H2O HNO2 + HO-
and you CANNOT use here Kw=Ka.Kb in this case.
B) A 0.001 aqueous solution of HClO4 (Ka = large) has a pOH of _______.
1) 13.999
2) 11
3) -3
4) -11
5) 3
Answer: It is a strong acid, therefore [acid] = [H3O+] = 0.001
pH = -log [H3O+] = 3
We know that pH + pOH = 14,
pOH = 14 – pH = 14 – 3 = 11
pf2

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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

  1. 2.2 10-
  2. 2.2 10-
  3. 2.2 10 3
  4. 4.5 10-
  5. None of the above

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 _______.

  1. 11
  2. 3

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.

  1. 1.0 10-
  2. 1.0 10-
  3. None of the above

Answer: carbonate is a weak base (Kb < 1). Therefore the [OH-] < [base]. We need to use an ICE table.

CO 3 2-^ + H 2 0HCO 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.