The calculation of pH is an important resource the student has to determine character. acidic, basic or neutral of a solution. In this text we will propose tips for calculating the pH of a buffer solution in a simple way.
It is noteworthy that a buffer solution can be formed by the following mixtures:
Mixture of a salt with a weak base, which must have the same cation as the salt. It is a basic buffer;
Mixing a salt with a weak acid, which must have the same anion as the salt. It is an acidic buffer.
Let's go to the tips?!
1st Tip: Formulas according to the type of buffer solution
When you have an acidic buffer solution, use:
pH = pKa + log [salt]
[acid]
When you have a basic buffer solution, use:
pOH = pKb + log [salt]
[base]
When you have a basic buffer solution and a different Kw (Water Ionization Constant), use:
pH = pKb - pKb - log [salt]
[base]
2nd Tip: When the exercise provides the concentrations of the participants and ionization constant...
We will have the concentration of the acid or base that forms the solution;
We will have the concentration of salt that forms the solution;
We will have the ionization constant (Ka or Kb) of the acid or base that forms the solution.
Example: (UNIFOR-CE-Adapted) A mixture of lactic acid (CH3CH(OH)COOH) and sodium lactate (CH3CH(OH)COONa) in aqueous solution works as a buffer solution, that is, it practically does not change its pH by the addition of H+ or oh-. A solution containing 0.12 mol/L of lactic acid and 0.12 mol/L of sodium lactate has a pH that can be calculated by the equation:
pH = pKa + log [salt]
[acid]
Ka = 1.0x10-4 = acid ionization constant. Neglecting the amount of acid that undergoes ionization, determine the pH value of the solution.
Resolution:
In this example, we have a buffer solution made up of salt and acid. The data provided are:
[salt] = 0.12 mol/L
[acid] = 0.12 mol/L
Ka = 1.10-4
NOTE: the exercise gave the Ka, but in the formula we use the pKa, which is simply – logKa.
As it is an acid buffer, just use the expression:
pH = pKa + log [salt]
[acid]
pH = - log 1.10-4 + log 0,12
0,12
pH = - log10-4 + log 0,12
0,12
pH = 4.log 10 + log 1
pH = 4.1 + 0
pH = 4
3rd Tip: When exercise requires changing the pH of a buffer solution that has received an amount of strong acid or base...
Exercise will provide the concentration of acid or base that forms it;
We will have the concentration of salt that forms the solution;
We will have the ionization constant (Ka or Kb) of the acid or base that forms the solution;
Exercise will provide the pH value of the buffer after adding the strong acid or base;
It is necessary to find the pH value of the buffer before adding the acid or strong base;
Then we must subtract the pH after addition from the pH before the addition.
Example: (Unimontes-MG) One liter of buffer solution contains 0.2 mol/L of sodium acetate and 0.2 mol/L of acetic acid. By adding sodium hydroxide, the pH of the solution changed to 4.94. Considering that the pKa of acetic acid is 4.76 at 25°C, what is the change in pH of the buffer solution?
Resolution: In this example we have a buffer solution formed by salt and acid. The data provided are:
pH after addition of strong base = 4.94
[salt] = 0.2 mol/L
[acid] = 0.2 mol/L
pKa = 4.76
Initially we must calculate the pH of the buffer before adding the strong base. For this, we must use the expression for acid buffer:
pH = pKa + log [salt]
[acid]
pH = 4.76 + log 0,2
0,2
pH = 4.76 + log 1
pH = 4.76 + 0
pH = 4.76
Finally, we subtract the pH after addition of base from the pH before addition:
ΔpH = after - before addition of base
ΔpH = 4.94 - 4.76
ΔpH = 0.18
4th Tip: Calculating the pH of a buffer when the exercise provides the mass of one of the participants
Exercise will provide the concentration or amount of matter of acid, base, or salt that forms it;
When the exercise provides the amount of matter (mol), it will also provide the volume, because in the pH calculation we use concentration (dividing the mol by the volume);
We will have the ionization constant (Ka or Kb) of the acid or base that forms the solution;
It is necessary to calculate the molar mass and the amount of matter of the participant that was given the mass in the exercise.
Example: (UFES - Adapted) A solution was prepared by adding 0.30 mol of acetic acid and 24.6 grams of sodium acetate in a sufficient quantity of water to complete 1.0 liter of solution. The CH system3COOH and CH3COONa constitutes a buffer solution in which this system is in equilibrium. Thus, determine the pH of the prepared solution. (Data: Ka = 1.8×10-5, log 1.8 = 0.26)
Resolution:
The data provided by the exercise were:
Ka = 1.8×10-5
log 1.8 = 0.26
Volume = 1L
Number of moles of acid 0.30 mole
-
As the volume is 1L, so [acid] = 0.30 mol/L
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Mass of salt used = 24.6 g
First: We must calculate the molar mass (M1) of salt:
CH3COONa
M1 = 1.12 + 3.1+ 1.12 + 1.16 + 1.16 + 1.23
M1 = 12 + 3 + 12 + 16 + 16 + 23
M1 = 82 g/mol
Second: Now let's determine the number of moles of salt by dividing the mass provided by the exercise by the molar mass found:
n = 24,6
82
n = 0.3 mol
Third: We must calculate the molar concentration of the salt by dividing the number of moles by the volume supplied:
M = no
V
M = 0,3
1
M = 0.3 mol/L
Room: We must calculate the pH using the expression for acidic buffer solution:
pH = pKa + log [salt]
[acid]
pH = -log 1.8.10-5 + log 0.3
0,3
pH = 5 - log 1.8 + log 1
pH = 5 - 0.26 + 0
pH = 4.74
5th Tip: Calculating the pH of a buffer solution that was prepared by mixing an acid and a base
We will have the molar concentration and volume of the acidic solution;
We will have the molar concentration and volume of the basic solution;
We will have the ionization constant of the acid or the base;
Determine the number of moles of acid and base used in the preparation (multiplying the molar concentration by the volume);
Respect the stoichiometric ratio, that is, for each H+ of the acid, an OH- of the base is used to neutralize;
As acid and base neutralize each other and form a salt, we must know if there is any acid (acid buffer) or base (basic buffer) left;
Determine the molar concentration of leftover and salt by dividing their mole numbers by the volume (sum of the volumes used in the preparation).
Example: (UEL) Buffer solutions are solutions that resist the change in pH when acids or bases are added or when dilution occurs. These solutions are particularly important in biochemical processes, as many biological systems are pH dependent. For example, the pH dependence on the cleavage rate of the amide bond of the amino acid trypsin by the enzyme is mentioned. chymotrypsin, in which a change in one unit of pH 8 (optimal pH) to 7 results in a 50% reduction in action enzymatic. For the buffer solution to have a significant buffering action, it must have comparable amounts of conjugated acid and base. In a Chemistry laboratory, a buffer solution was prepared by mixing 0.50 L of ethanoic acid (CH3COOH) 0.20 mol L-1 with 0.50 L of sodium hydroxide (NaOH) 0.10 mol L-1. (Given: pKa of ethanoic acid = 4.75)
Resolution:
The data provided by the exercise are:
[acid] = 0.20 mol/L
Acid volume = 0.5 L
[base] = 0.10 mol/L
Base volume = 0.5 L
pKa = 4.75
First: calculation of the number of moles of the acid (na):
na = 0.20. 0,5
na = 0.1 mol
Second: calculation of the number of moles of the base:
nb = 0.10. 0,5
nb = 0.05 mol
Third: Determine who is left in the solution:
Ethanoic acid has only one ionizable hydrogen, and the base has a hydroxyl group, so the ratio between them is 1:1. So the number of moles of both should be the same, but we have a larger amount (0.1 mole) of acid than the amount of base (0.05 mole), leaving 0.05 mole of acid.
Room: Determination of the number of moles of the salt
As the amount of salt formed is always related to the components of smaller stoichiometric proportion (balancing), in this example, the amount of salt follows coefficient 1, that is, its mole number is also 0.5 mol.
Fifth: Determination of molar concentration of acid and salt
0.5 L of acid was mixed with 0.5 L of base, resulting in a volume of 1 L. Thus, the acid and salt concentration are equal to 0.05 mol/L.
Sixth: pH determination
As the buffer is acidic, just use the values in the following expression:
pH = pKa + log [salt]
[acid]
pH = 4.75 + log 0,05
0,05
pH = 4.75 + log 1
pH = 4.75 + 0
pH = 4.75
6th Tip: When the exercise questions the new pH value after adding an amount of strong acid or base...
We will have the value of the molar concentration of the acid or base that was added to the buffer;
We must have the molar concentration of the salt, acid or base that forms the buffer. In case we don't have it, just calculate it as seen in the previous tips;
The added concentration will always be subtracted from the acid or base concentration;
The added concentration will always be added to the salt concentration.
Example: Determine the pH of the buffer solution after adding 0.01 mol of NaOH knowing that, in 1.0 L of the prepared solution, we have 0.05 mol/L of acetic acid and 0.05 mol/L of sodium acetate. Data: (pKa = 4.75, log 0.0666 = 0.1765)
Resolution:
Data provided:
[salt] = 0.05 mol/L
[acid] = 0.05 mol/L
[base added to buffer] = 0.01 mol/L
pKa = 4.75
pH = pKa – log (salt - base)
(acid + base)
pH = 4.75 - log (0,05 - 0,01)
(0,05 + 0,01)
pH = 4.75 - log 0,04
0,06
pH = 4.75 - log 0.666
pH = 4.75 + 0.1765
pH = 4.9265
By Me. Diogo Lopes Dias
