A buffer solution is made that is 0.353 M in CH3COOH and 0.353 M in CH3COO- . (1) If Ka for CH3COOH is 1.80×10-5 , what is the pH of the buffer solution? (2) Write the net ionic equation for the reaction that occurs when 0.090 mol HI is added to 1.00 L of the buffer solution. Use H3O+ instead of H+ .

(1) pH of the Buffer solution = 4.82

(2) pH of Buffer solution after addition of HI = 4.6

Explanation:

(1) A buffer solution consisting a mixture of a weak acid and its conjugate base is termed as Acidic Buffer.

CH₃COOH is a weak acid and its conjugate base CH₃COO⁻ is strong.

Ka (acid dissociation constant) = 1.80 x 10⁻⁵

⇒pKa = -Log(1.80 x 10⁻⁵ )

⇒pKa = 4.82

Using Henderson equation

pH = pKa + Log

pH = 4.82 + Log

pH = 4.82

(2) HI is a strong acid and dissociate completely.

  HI ⇆ H⁺ + I⁻

Due to its complete ionization a common ion effect arrises .

Assume all acid react with acetate ion to produce acetic acid (CH₃COOH)

CH₃COO⁻ + H₃O⁺ ⇄ CH₃COOH + H₂O

[H₃O⁺] added =  1 lit 0.09 mole = 0.09 molar

[CH₃COOH] = 0.353 + 0.09 = 0.443

[CH₃COO⁻]  = 0.353 - 0.09 = 0.263

Using Henderson equation

pH = pKa + Log

pH = 4.82 + Log

pH = 4.82 - .22 = 4.6

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

answer; economic order quantity;