Solved Examples on Buffer Solution
Example 1: What is the pH of a buffered solution of 1.5 M NH3 and 2.5 M NH4Cl when 0.5 M HCl is added to the solution?
Solution:
We know that,
pKb of ammonia is 4.75
pKa = 14 – pKb.
= 14 – 4.75 = 9.25
Now, on adding 0.5 M HCl
0.5 M H+ ions are available in the aqueous solution which reacts with 0.5 M NH3 to form 0.5 M NH4Cl
Now the remaining concentration of ammonia is 1 M and that of NH4Cl is 3 M.
Using the Henderson-Hasselbalch equation,
pKa – log ([salt]/[acid]) = 9.25 – log (3/1)
= 9.25 – 0.477
= 8.773
Example 2: How many moles of sodium formate and formic acid are required to prepare 1.00 L of a 0.25 mol/L buffer solution with pH 4.00? (given pKa of Sodium formate is 3.86)
Solution:
pH = pKa + log([A−][HA])
Given, pKa of Sodium formate is 3.86
6.00 = 3.86 + log([A−][HA])
log ([A−][HA]) = 4 – 3.86
log ([A−][HA]) = 0.14
[A−][HA] = 100.14
= 1.38
[A⁻] = 1.38[HA]
Given,
[A⁻] + [HA] = 0.25 mol/L
1.38[HA] + [HA] = 0.25 mol/L
2.38[HA] = 0.25 mol/L
[HA] = 0.25 mol/L / 2.38
[HA] = 0.105 mol / L
[A⁻] = (0.250 – 0.105) mol/L
= 0.145 mol/L
Thus, 0.105 mol of sodium formate and 0.145 mol of formic acid is required to form the required buffer solution.
Buffer Solution
Buffer Solution is a special aqueous solution that resists the change in its pH when some quantity of acid and Base is added. Many fluids, such as blood, have specific pH values of 7.14, and variations in these values indicate that the body is malfunctioning. The change in pH of Buffer Solutions on adding a small quantity of acid or bases is very minimal and hence they are used to make solutions that resist the change in pH.
Let us learn about Buffer solution, its types, and others in this article.
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