1.)An ammonia/ammonium buffer solution contains 0.35 M NH3 and 0.72 M NH4+. The Kb value of ammonia is 1.8×10−5. Calculate the pH of this buffer.
2.) Nitrous acid has a Ka of 4.5×10−4. What is the pH of a buffer solution containing 0.15 M HNO2 and 0.20 M NO−2?
3.) Acetic acid has a Ka of 1.8×10−5.
What is the pH of a buffer solution containing 0.14 M HC2H3O2 (acetic acid) and 0.20 M C2H3O2−?
To calculate the pH of a buffer solution, you need to use the Henderson-Hasselbalch equation, which relates the pH of the solution to the concentrations of the acid and its conjugate base.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([base]/[acid])
1.) For the ammonia/ammonium buffer solution, the ammonia (NH3) acts as the base and ammonium (NH4+) acts as the acid. The pKa of ammonia can be calculated using the Kb value using the equation pKa + pKb = 14. Since the Kb value of ammonia is 1.8×10−5, the pKa can be calculated as 14 - pKb = 14 - (-log[1.8×10−5]) = 14 + 4.75 = 18.75.
Using the Henderson-Hasselbalch equation:
pH = 18.75 + log([NH4+]/[NH3])
= 18.75 + log(0.72/0.35)
= 18.75 + log(2.057)
= 18.75 + 0.313
= 19.063
Therefore, the pH of the ammonia/ammonium buffer solution is approximately 19.063.
2.) For the nitrous acid buffer solution, nitrous acid (HNO2) acts as the acid and nitrite ion (NO−2) acts as the conjugate base.
Using the Henderson-Hasselbalch equation:
pH = pKa + log([NO−2]/[HNO2])
= -log(4.5×10−4) + log(0.20/0.15)
Since log([NO−2]/[HNO2]) = log(0.20/0.15) is a ratio that is less than 1, you can rewrite the equation as:
pH ≈ pKa + log([NO−2]/[HNO2])
For calculations, you can use the approximation given that the ratio is small. In this case, the pH would be approximately equal to the pKa of nitrous acid: pH ≈ -log(4.5×10−4) = 3.35.
Therefore, the pH of the nitrous acid buffer solution is approximately 3.35.
3.) For the acetic acid buffer solution, acetic acid (HC2H3O2) acts as the acid and acetate ion (C2H3O2−) acts as the conjugate base.
Using the Henderson-Hasselbalch equation:
pH = pKa + log([C2H3O2−]/[HC2H3O2])
= -log(1.8×10−5) + log(0.20/0.14)
Since log([C2H3O2−]/[HC2H3O2]) = log(0.20/0.14) is a ratio that is greater than 1, you can rewrite the equation as:
pH ≈ pKa - log([HC2H3O2]/[C2H3O2−])
For calculations, you can use the approximation given that the ratio is large. In this case, the pH would be approximately equal to the pKa of acetic acid: pH ≈ -log(1.8×10−5) = 4.74.
Therefore, the pH of the acetic acid buffer solution is approximately 4.74.
1. Use the Henderson-Hasselbalch equation. KaKb = Kw and you get Ka from that and pKa follows.
2. Ditto for 2.
3. Ditto for 3.