# Answered Essay: Calculate the pH of the following solutions using the Henderson-Hasselbalch equation: a.

Calculate the pH of the following solutions using the Henderson-Hasselbalch equation: a. 0.03M acetic acid and 0.03M sodium acetate b. 0.02M acetic acid and 0.04M sodium acetate c. 0.04M acetic acid and 0.02M sodium acetate Draw the D-form of the amino acid alanine. 8. What is the ratio of lactate/lactic acid in blood if the pH is 7.0? (look at page 30 in text) What is the ratio of acetate to acetic acid in a solution where the pH is 4.5? Also what is the % of the buffer in the acetate form at this pH? Write out (draw structures) for the dissociation steps for aspartic acid, starting out with the low pH form. This will be three separate steps. Draw the predominant form (i.e. > 50%) for each of the amino acids their structures, you do not need to do calculations. Just compare the pH to each of the ionizable groups (i.e. pK_1, pK_2, and pK_R) to decide if the proton is on or off these groups below. a. glycine at pH 12 b. lysine at pH 7.0 c. alanine at pH 2.0 d. aspartic acid at pH 12

Question 6.

The Buffer Equation (Henderson Hasselbach) ACIDIC

This is an acidic buffer; since there is a weak acid + conjugate base:

A buffer is any type of substance that will resist pH change when H+ or OH- is added.

This is typically achieved with equilibrium equations. Both type of buffer will resist both type of additions.

When a weak acid and its conjugate base are added, they will form a buffer

The equations that explain this phenomena are given below:

The Weak acid equilibrium:

HA(aq) <-> H+(aq) + A-(aq)

Weak acid = HA(aq)

Conjugate base = A-(aq)

Neutralization of H+ ions:

A-(aq) + H+(aq) <-> HA(aq); in this case, HA is formed, H+ is neutralized as well as A-, the conjugate

Neutralization of OH- ions:

HA(aq) + OH-(aq) <-> H2O(l) + A-(aq) ; in this case; A- is formed, OH- is neutralized as well as HA.

Note that the equilibirum equation can be mathematically manipulated in order to favour the “buffer” construction

Recall that, in equilibrium

Ka = [H+][A-]/[HA]

Multiply both sides by [HA]

[HA]*Ka = [H+][A-]

take the log(X)

log([HA]*Ka) = log([H+][A-])

log can be separated

log([HA]) + log(Ka) = log([H+]) + log([A-])

note that if we use “pKx” we can get:

pKa = -log(Ka) and pH = -log([H+])

substitute

log([HA]) + log(Ka) = log([H+]) + log([A-])

log([HA]) + -pKa = -pH + log([A-])

manipulate:

pH = pKa + log([A-]) – log([HA])

join logs:

pH = pKa + log([A-]/[HA])

which is Henderson hasselbach equations.

Now,

substitute data:

for acetic acid, pKa = 4.75

pH = 4.75 + log(acetate/acetic acid)

a)

pH = 4.75 + log(0.03/0.03)

pH = 4.75

b)

pH = 4.75 + log(0.04/0.02)

pH = 5.05102

c)

pH = 4.75 + log(0.02/0.04)

pH = 4.4489

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