pH and Buffers

Noteworthy

Always keep in mind the following equations:

1) pH=-log[H₃O⁺];

2) pOH=-log[OH^-]

3) K_a=[conjugate base][H₃O⁺]/[conjugate acid];

4) K_b=[conjugate acid][OH^-]/[conjugate base]

5) pK_a=-logK_a;

6) pK_b=-logK_b

7) Just for water pK_w=pH⁺+ pOH; K_w=[H₃O⁺] [OH^-] ;

8) For a conjugate acid/base solution would be pK_w=pK_a+ pK_b; K_w = K_a X K_b

9) pH=pK_a + log [conjugate base]/[conjugate acid] (Henderson-Hasselback equation)

10) moles = mass (grams)/molecular weight (daltons; grams/mole)

11) Molar concentration (M) = moles/liter

12) Dilution factors of a stock solution could be calculated using this formula: C₁V₁=C₂V₂ ; where C is concentration and V is volume of the solution.

Problem 1 (pH)

What would the pH of a 0.2 M acetic acid solution be? acetic acid pKa = 4.75

First, check what data you have and what data you need

I have: pKa =4.75 and [conjugate acid] = 0.2 M

I need to calculate: pH = ?

To calculate pH I must used the first equation pH=-log[H₃O⁺], however I do not the value of [H₃O⁺].

The third equation could be use to calculate the value [H₃O⁺]. This equation uses two values that I know, or that I can calculate, K_a and [conjugate acid]

K_a can be calculated by rewriting the fifth equation; K_a=10^-pKa

Adding values I got: K_a=10^-4.75 = 1.78 X 10^-5

Adding values to the third equation I got; 1.78 X 10^-5 = [conjugate base][H⁺]/(0.2 M-[conjugate base]);

Remember when a weak acid dissociates a proton at a time usually produces this:

[conjugate base] = [H₃O⁺] ; (HA ---- A^- + H⁺).

This is the case for acetic acid; CH₃COOH (conjugate acid) + H₂O -------- CH₃COO^- (conjugate base) + H₃O⁺

So, when one molecule of conjugate acid dissociates one molecule of a hydronium ion and one molecule of conjugate base are produced. This in fact reduces the concentration of the conjugate acid! Because is transformed in its corresponding conjugate base.

Rewriting this equation 1.78 X 10^-5 = [conjugate base][H⁺]/(0.2 M-[conjugate base]; I got: 1.78 X 10^-5 = [H⁺]²/(0.2 M-[H⁺]) ;

Resolving the previous equation for [H⁺] I got:

[H⁺]² = 1.78 X 10^-5 X 0.2 M - 1.78 X 10^-5 [H⁺]

or

[H⁺]² + 1.78 X 10^-5 [H⁺] - 0.356 X 10^-5 M = 0;

This last equation in mathematics is known as quadratic equation

aX² + bX + c = 0

Check these links:

https://www.mathsisfun.com/algebra/quadratic-equation.html

http://www.biology.arizona.edu/biomath/tutorials/quadratic/QuadraticFunctions.html

To resolve a quadratic equation I would use the quadratic formula: X = (-b +/- √b²- 4ac)/2a

Then when I plugged in the data I got: X = (-1.78 X 10^-5 +/- √3.17 X 10^-10 - (-1.42 X 10^-5)) / 2 = 1.9 X 10^-3

Remember X = [H⁺] needs to be a positive value; also the final number should not be equal or bigger than the concentration of the acid, it would not make any sense to get equal values or more protons than the starting number of acid to be dissociated.

Now that I know that [H⁺] = 1.9 X 10^-3 M; then I can use the equation: pH =-log[H⁺].

Plugging in the numbers I got that the result is: pH = -log (1.9 X 10^-3) = 2.72

Interesting fact of this solution: just (1.9 X10^-3/0.2) X 100% = 0.95% of the total acetic acid molecules dissociate in a water solution. This makes acetic acid a weak acid!

Problem 2 (pH)

What would the pH of a 0.1 M Tris-base solution be? pK_a = 8.07

First, check what data you have and what data you need

I have: pKa =8.07 and [conjugate base] = 0.1 M

I need to calculate: pH = ?

Tris(hydroxymethyl)aminomethane also known as Tris, has weak acid group (I know this because there is just one pK_a value). Check the following link for the structure and function of Tris: https://www.labome.com/method/Protein-Purification.html

Tris(NH₂)base is the conjugate base of Tris(NH₃⁺)acid. In solution Tris-base would do the following:

Tris(NH₂)base + H₂O ------------ Tris(NH₃⁺)acid + OH^-

In this case I have to use the fourth equation from my top list:

K_b (constant of association) = [Tris(NH₃⁺)acid] [OH^-] / [Tris(NH₂)base].

Considering that for each reacting molecule of tris-base one molecule of tris-acid and one molecule of hydroxyl ion would be produced I could said that [Tris(NH₃⁺)acid] = [OH^-]. Also, remember the initial concentration of Tris base would be reduced every time than produces its conjugate acid until reaching equilibrium ([Tris(NH₂)base] - [Tris(NH₃⁺)acid]). Then, rearranging the last equation I got:

[OH^-] [Tris(NH₃⁺)acid] = K_b X ([Tris(NH₂)base] - [Tris(NH₃⁺)acid]) or [OH^-]² = K_b X ([Tris(NH₂)base] - [OH^-]) =

[OH^-]² = K_b [Tris(NH₂)base] - K_b[OH^-] a quadratic equation again

This equation would be used to calculate [OH^-], which would later be used to calculate [H₃O⁺] by using the seventh formula of my list.

First I need to calculate the K_b value using the eight equation of my list K_w=K_a X K_b.

Rewriting this equation I got K_b = K_w / K_a

K_a can be calculated by rewriting the fifth equation; K_a=10^-pKa

Adding values I got: K_a=10^-8.07 = 8.5 X 10^-9

Plugging then some values I calculate that K_b = 1 X 10^-14 (water dissociation constant) / 8.5 X 10^-9 = 1.2 X 10^-5

Adding the data into the quadratic equation I got:

[OH^-]² = 1.2 X 10^-5 (0.1M) - 1.2 X 10^-5 [OH^-];

or

[OH^-]² + 1.2 X 10^-5 [OH^-] -1.2 X 10^-6 = 0

To resolve a quadratic equation I would use the quadratic formula: X = (-b +/- √b²- 4ac)/2a

Then when I plugged in the data I got: X = (-1.2 X 10^-5 +/- √1.44 X 10^-10 - (-4.8 X 10^-5)) / 2 = 3.4 X 10^-3

Remember X = [OH^-] needs to be a positive value; also the final number should not be equal or bigger than the concentration of the base, it would not make any sense to get equal values or more hydroxyl ions than the starting number of base to be reacted.

Now that I know that [OH-] = 3.4 X 10^-3 M; then I can use the equations: pOH =-log[OH^-] and pK_w = pH + pOH to calculate pH.

Plugging in the numbers I got that the result is: pOH = -log (3.4 X 10^-3) = 2.47 and pH = 14 - 2.47 = 11.53

Interesting fact of this solution: just (3.4 X10^-3/0.1) X 100% = 3.4% of the total Tris base molecules reacts with water. This makes Tris base a weak base!

Problem 3 (Making buffers)

Explain how would you proceed doing a 50 ml solution of a 0.1 M Tris buffer pH 7.5? pK_a = 8.07,

Tris molecular weight = 121 gr/mole. You are given a 1 M solution of each acid and conjugate base.

Remember buffers always used as a mathematical principle the Henderson-Hasselbach equation (ninth equation).

pH=pK_a + log [conjugate base]/[conjugate acid]

Then, to do this solution I need to know:

1) The pH to be fixed, in this case is 7.5;

2) The pK_a of the acid, in this case is 8.07;

With the previous values and the equation then I can calculate the amount of acid and its conjugate base that should be added to the solution in order to generate this buffer.

Rewriting the HH formula I got that:

[conjugate base]/[conjugate acid]= 10^pH-pKa

Adding values

[Tris(NH₂)-base]/[Tris(NH₃⁺)-acid] = 10^7.5-8.07 = 0.27

Interesting fact: O.27 indicates that to get a pH lower than the pK_a of this acid, the amount of Tris(NH₃⁺)-acid should be almost four times higher than the amount of Tris(NH₂)-base.

Still I have not calculate yet the amount of each conjugate acid and base to be used. To do that I need to understand that the total amount of this buffer is made by the addition of both acid and conjugate base. Therefore I can make the following equation:

[Buffer] = [conjugate acid] + [conjugate base]; or moles of buffer = moles of conjugate acid + moles of conjugate base

In the case of this problem:

0.1 M of Tris buffer = [Tris(NH₂)-base] + [Tris(NH₃⁺)-acid]

If I rewrite my previous equation

[Tris(NH₂)-base] = 0.27 X [Tris(NH₃⁺)-acid]

I could substitute this value on the top equation, eliminating one variable (THANKS MATH, YOU ARE VERY USEFUL)

0.1 M of Tris buffer = 0.27 X [Tris(NH₃⁺)-acid] + [Tris(NH₃⁺)-acid]

Resolving the equation I got

0.1 M of Tris buffer / 1.27 = [Tris(NH₃⁺)-acid] = 0.078 M

To get the conjugate base concentration I just have to add my new value to the following equation

[Tris(NH₂)-base] = 0.27 X 0.078M = 0.022 M

These last two data indicates that to make 50 ml of the 0.1 M Tris buffer solution pH 7.5, I need to mix enough acid to make a 0.078 M solution in 50 ml and in the same solution have mixed 0.022 M of the conjugate base.

Because I am producing a 0.1M solution from 1M stock I have to use the twelve equation

C₁V₁=C₂V₂

Adding values for each component I got

Acid

1M V₁ = 0.078M (50 ml); V₁ = 3.9 M mL /1 M = 3.9 ml

Base

1M V₁ = 0.022M (50 ml); V₁ = 1.1 M mL /1 M = 1.1 ml

To do this solution I have to mix 3.9 ml of the 1 M Tris acid solution with 1.1 ml of the 1 M Tris base solution and then add 45 ml of water to get 50 ml of the 0.1 M Tris buffer solution pH 7.5

Now try with these problems:

Practice pH and Buffer problems.pdf