Question

Explain why it is not possible to prepare a buffer with a \(\mathrm{pH}\) of 6.50 by mixing \(\mathrm{NH}_{3}\) and \(\mathrm{NH}_{4} \mathrm{Cl}\).

Short answer

It is not possible to prepare a buffer with a pH of 6.50 by mixing NH₃ and NH₄Cl because the required ratio of concentrations for NH₃ and NH₄⁺, as calculated using the Henderson-Hasselbalch equation, would be approximately 0.00177, which indicates that the concentration of NH₃ must be much lower than the concentration of NH₄⁺. Since NH₃ is a weak base, having a solution predominantly composed of the base with only a small amount of the acidic NH₄⁺ would not likely result in an acidic solution (pH < 7).

Step-by-step solution

Recalling the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is given by: pH = pKa + log([A⁻]/[HA]). In our case, the weak acid HA is NH₄⁺ (from NH₄Cl) and the conjugate base A⁻ is NH₃.

Determining the pKa of NH₄⁺

The pKa of NH₄⁺ can be found from the Ka value of NH₄⁺. The Ka value is 5.56 x 10⁻¹⁰, so pKa = -log(Ka) = -log(5.56 x 10⁻¹⁰) = 9.255

Plugging the given pH value into the Henderson-Hasselbalch equation

We know the desired pH is 6.50 and we have the value of pKa. Now, we can plug these values into the Henderson-Hasselbalch equation to find the required ratio of [A⁻]/[HA]: 6.50 = 9.255 + log([NH₃]/[NH₄⁺])

Solving for the [NH₃]/[NH₄⁺] ratio

Subtract the pKa from both sides of the equation: -2.755 = log([NH₃]/[NH₄⁺]). To find the [NH₃]/[NH₄⁺] ratio, we can take the antilog (10^x) of both sides of the equation: 10^(-2.755) = [NH₃]/[NH₄⁺]. Thus, [NH₃]/[NH₄⁺] ≈ 0.00177.

Analyzing the calculated ratio

The calculated ratio of [NH₃]/[NH₄⁺] tells us that the concentration of NH₃ must be much lower than the concentration of NH₄⁺ in order to achieve a pH of 6.50. Considering that NH₃ is a weak base, it is unlikely that the solution would be acidic (pH < 7) if it was predominantly composed of the base with only a small amount of the acidic NH₄⁺. Therefore, it is not possible to prepare a buffer with a pH of 6.50 by mixing NH₃ and NH₄Cl.