Calculating the pH of an Acid/Base Buffer System

Begin with 300mL of 0.200M Acetic Acid and 256mL 0.140M Potassium Hydroxide.

a.) Calculate the Initial pH of CH₃COOH:

b.) Calculate the pH after adding 256mL of KOH:

Well, we might start with some discussion about pH.  pH is a scale used to describe the strength of an acid or a base.  The scale ranges from 0 to 14.  Acidity increases as pH aproaches zero.  Basicity increases as pH approaches 14.  Strong acids generally have a pH below 2, strong bases generally have a pH above 12.  pH = -log₁₀[H₃O⁺].¹  Hydronium is what makes acids acids.  It is just water with that extra Hydrogen cation.  It is impossible to have pure Hydronium, however the strongest acids in aqueos solution produce a very large amount of hydronium.  Some other important pieces here are the k_a value and for our buffer system the pk_a value.  K_a is a constant value created by the ratio of acid and its conjugate base multiplied by the concentration of Hydronium cations.  It should be noted that the concentration of Hydronium cations and the Conjugate base is equal.  k_a = [H₃O⁺][Acid^-] / [Acid].  Also..  pk_a = -log₁₀k_a.  There is also a k_b value which is a constant discribed similarily by the ratio of a base and its conjugate acid multiplied by the concentration of Hydroxide anions.  Also k_w = 1.0E-14.  k_a x k_b = k_w.  For the reaction that we are discussing, a weak acid being reacted with a strong base, there is an equation called the Henderson/Hassleboff equation.  This is used to calculate pH in an acid/base buffer system.  We know to use this equation when there are both acid and conjugate base present.  When either one becomes completely used up, the equation is no longer of use to us.   pH = pK_a + log₁₀([base] / [acid]).

Another equation that we will find useful is the Molarity ('M') equation:  M = n/V_L or Molarity = number of mols divided by the volume of the solution in Litres. 

a.)

Step 1 is to look at our acid.  Acetic Acid (CH₃COOH).  The K_a value for this acid is. 1.8197E-5.  We will use this in the relationship described above to find the concentration of Hydronium cations.

The simple dissociation of Acetic Acid in Water is written as:

CH₃COOH + H₂O --> H₃O⁺ + CH₃COO^-.

So the relation ship works as  (k_a x [CH₃COOH])^1/2 = [H₃O⁺].

This piece of course assumes 1. that we can truncate the subtraction of Hydronium cation concentration from the Acetic Acid in the denominator of our original relationship because the amount would be insignificant to the overall precision of our equation. and 2. that the error between the known pH and the calculated pH at standard conditions is less than 3%. I will not take time to prove the assumptions.

so ((1.8197E-5)(0.200M))^1/2 ≈ 3.6394E-6  This is our initial concentration of Hydronuim cations in our aqueous solution.

-log₁₀3.6394E-6 = 5.44  This is the pH value of our Acetic Acid defining it as a weak acid.

b.)

Step 1 is to write out our equation for the buffer system.

CH₃COOH + KOH --> CH₃COO^-K⁺ + H₂O.

Upon reviewing the equation we can see that it is a balanced equation.  Therefore the production of conjugate base (Potassium Acetate)² is a 1 to 1 relationship with the loss of acid (Acetic Acid).  Also the KOH we add (a strong base) gets used completely.  If it did not get used completely the overwhelming strong base would destroy the buffer system.

Step 2 is to do some math about molarity and number of moles used and made of the different chemicals in our relationship.

We need to find out how many moles of each substance we have.

(300E-3L)(0.200M) = 0.060 moles CH₃COOH

(256E-3L)(0.140M) = 0.036 moles KOH

Now in our one to one relationship, for every mole of strong base (KOH) added we create one mole of our conjugate base (CH₃COO^-K⁺) therefore we need only to subtract our added strong base from our weak acid and we will then have the amount of acid used, and also the amount of conjugate base created.

0.060 - 0.036 = 0.024

we have leftover:
0.024 moles of Acetic Acid
0.000 moles of Potassium Hydroxide

we have made:
0.036 moles of Potassium Acetate

into our equation  pH = pK_a + log₁₀([base] / [acid])

because the amount of conjugate base we have is greater than the leftover weak acid, the number we produce will be a pOH which is a scale that also describes the basicity/acidity of a solution.  We will need to convert back to pH by subtracting the pOH value from 14.0

pOH = 4.74 + log₁₀(0.036 / 0.024) ≈ 4.92

pH = 14.0 - 4.92 = 9.08

we can skip converting our molar values for the weak acid and conjugate base into our new volume (556ml) solution because the ratio inside the logarithm would not change.

you will notice that the new pH is considerably higher than the initial pH.

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After Discussion:

I now ask you:

a.) What would the pH have been if we had only added 47ml of the strong base initially.

b.) How many ml KOH will we need to add to our initial solution reach equivalence?

c.) What important relationship built into the Henderson/Hassleboff equation do we see when the concentration of our weak acid and our conjugate base are equal? 

bonus.) Which 2nd derivative calculus property might let us find the pH of the equivalence point of our buffer more precisely?

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1. it is understood that brackets represent concentration. ie.) [base] = the concentration of the base.
2. this is known as the potassium salt of the conjugate base of acetic acid.  the potassium has no effect on the pH of the solution for our purposes.

©2005 Joseph D. Gamble (student) 10/25/2005