Types of buffer solutions
Maintaining the pH of body fluids so that the excretion of H + ions in the kidneys is not disturbed, namely dihydrogen posphat acid (H2PO4-) with a base monohydrogen posphat (HPO42-)
Maintain the pH of processed foods in cans so they are not easily damaged / oxidized (benzoic acid with sodium benzoate).
In addition, the application of this buffer solution can be found in everyday life such as eye drops.
A buffer solution can be formed from a mixture of weak acids and their conjugate bases or weak bases and conjugate acids. Based on the constituent acid base, the buffer solution can be divided into 2, which are as follows:
1. Acid buffer solution
Acid buffer solution is a buffer solution formed from weak acids and their conjugate bases. Acid buffer solutions have a pH of less than 7.
Example: CH₃COOH (weak acid) and CH₃COO– (conjugate base).
2. Base buffer solution
Base buffer solution is a buffer solution formed from a weak base and conjugate acid. Base buffer solution with a pH greater than 7.
Example: NH₃ (weak base) and NH₄ + (conjugate acid).
Example Solution / Buffer Buffer
Example of a Buffer Buffer Solution
Determine the pH of the solution if 800 ml of 0.1 M CH3COOH solution mixed with 400 ml of 0.1 M CH3COONa solution (Ka CH 3 COOH = 1.8 × 10-5)!
Determine the pH of the solution if 400 ml of 0.5M NH4OH solution is mixed with 100 ml of 0.5M NH4Cl solution (Kb NH4OH = 1.8 × 10-5)
A total of 50 ml of solution consisting of 1M CH3COOH and 1M CH3COONa were added with 1M HCl solution. Determine the pH of the solution after adding 1M HCl! (Ka = 1.8 x 10-5)
A total of 50 ml of solution consisting of CH3COOH 1M and CH3COONa 1M plus 50 ml of water. Determine the pH of the solution after dilution!
Answer Buffer Buffer Solution
Answer No. 1
mol CH3COOH = 800 x 0.1 = 80 mmol
mol CH3COONa = 400 x 0.1 = 40 mmol
[H +] = Ka.na / nbk
= 1.8 x 10-5 x (80/40)
= 3.6 x 10 -5
pH = -log 3.6 x 10 -5
= 5 - log 3,6
Answer No. 2
NH3 mole = 400 x 0.5 = 200 mmol
NH4Cl mole = 100 x 0.5 = 50 mmol
[OH–] = 1.8 x10 -5 x (200/50)
= 7.2 x 10 -5
pOH = log 7.2 x 10 -5
= 5 - log 7.2
pH = 14 - (5-log 7.2)
= 9 + log 7.2
Answer No. 3
mol CH3COOH = 50 x 1 = 50 mmol
mol CH3COONa = 50 x 1 = 50 mmol
mole HCl = 1 x 1 = 1 mmol
CH3COONa + HCl —-> CH3COOH + NaCl
At first: ……………… 50 mmol ……… .. 1 mmol …… .50 mmol -
React: …………………. 1 mmol ……… .. 1 mmol …… 1 mmol …… ..1 mmol
Remaining ………: ……………… 49 mmol …………. - ………… ..51 mmol …… .1 mmol
So the pH = -log (1.8 x 10-5 x 51/49)
= -log 1.87 x 10-5 = 5 - log 1.87
Answer No. 4
CH3COOH dilution: V1.M1 = V2.M
50 × 1 = 100xM2
M2 = 0.5
CH3COONa Dilution: V1.M1 = V2.M2
50 × 1 = 100xM2
M2 = 0.5