50 ml of 0.1 m naoh is added to 75 ml of 0.1 m nh4 cl to make a basic buffer. If pka of nh4 is 9.26, calculate ph.
Answers
pH of the solution is 9.73
Given-
- Volume of NaOH = 50 ml
- Concentration of NaOH = 0.1 M
- Volume of NH₄Cl = 75 ml
- Concentration of NH₄Cl = 0.1 M
- pKa of NH₄Cl = 9.26
- Kb of NH₃ = 1.8 × 10⁻⁵
The given reaction will be
NH₄Cl + NaOH → NaCl + NH₄OH
Initial mole of NH₄Cl = 50 mL x 0.2 mole/L = 10 m moles
Initial mole of NaOH = 75 mL x 0.1 mole/L = 7.5 m moles
After mixing concentration of NaOH will be zero because it has been completely reacted.
Reacted concentration of NH₄Cl is equals to the initial concentration of NaOH. So,
[NH₄Cl] reacted = [NaOH] initial = 7.5 m mole
[NH₄Cl] final = [NH₄Cl] initial - [NH₄Cl] reacted = 10 m moles - 7.5 m moles = 2.5 m moles
[NH₄OH] = [NH₄Cl] reacted = 7.5 m moles
Volume of final solution will be = 0.5 L+ 0.5 L= 1 L
pH can be calculated by-
pH = 14 - pKb NH₃ - log [NH₄⁺]/[NH₃]
by substituting the values we get
pH = 14 + log 1.8 × 10⁻⁵ - log 2.5/7.5
pH = 9.73
Explanation:
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