Question

You want to make a buffer with a pH of 10.00 from \(\mathrm{NH}_{4}^{+} / \mathrm{NH}_{3}\) (a) What must the \(\left[\mathrm{NH}_{4}{ }^{+}\right] /\left[\mathrm{NH}_{3}\right]\) ratio be? (b) How many moles of \(\mathrm{NH}_{4}\) Cl must be added to \(465 \mathrm{~mL}\) of an aqueous solution of \(1.24 \mathrm{M} \mathrm{NH}_{3}\) to give this pH? (c) How many milliliters of \(0.236 \mathrm{M} \mathrm{NH}_{3}\) must be added to \(2.08 \mathrm{~g}\) of \(\mathrm{NH}_{4} \mathrm{Cl}\) to give this \(\mathrm{pH}\) ? (d) What volume of \(0.499 \mathrm{M} \mathrm{NH}_{3}\) must be added to \(395 \mathrm{~mL}\) of \(0.109 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) to give this \(\mathrm{pH}\) ?

Short answer

The required ratio of [NH4+] / [NH3] is approximately 5.62 to create a buffer solution with a pH of 10.00.

Step-by-step solution

Calculate the ratio of [NH4+] / [NH3] using the Henderson-Hasselbalch equation

Since the desired pH value is 10.00, we can use the Henderson-Hasselbalch equation to find the required ratio of [NH4+] / [NH3]: \(pH = pKa + \log\frac{[NH4+]}{[NH3]}\) First, we need to find the pKa, which is the negative logarithm of Ka value, and then we can plug in the known values and solve for the ratio: \(pKa = -\log(5.56\times10^{-10})\) \(pKa = 9.25\) Now, substitute pH and pKa in the Henderson-Hasselbalch equation and solve for the ratio: \(10.00 = 9.25 + \log\frac{[NH4+]}{[NH3]}\) \(\log\frac{[NH4+]}{[NH3]} = 0.75\) Finally, take the antilog (base 10 exponent) of both sides to find the ratio of [NH4+] / [NH3]: \(\frac{[NH4+]}{[NH3]} = 10^{0.75} \approx 5.62\) Thus, the ratio of [NH4+] / [NH3] must be approximately 5.62.

Determine the moles of NH4Cl needed (Part b)

The volume of NH3 solution given is 465 mL, and its concentration is 1.24 M. From this information, we can determine the moles of NH3: moles of NH3 = volume * concentration = (465 mL)(1.24 M) = 576.6 mmol Now, we need to find the moles of NH4+ needed to achieve the desired ratio of 5.62. Since the ratio is \(\frac{[NH4+]}{[NH3]}\), the equation becomes: moles of NH4+ = moles of NH3 * desired ratio moles of NH4+ = (576.6 mmol)(5.62) ≈ 3240 mmol of NH4+ As NH4Cl contains one NH4+ ion per molecule, 3240 mmol of NH4Cl is needed to be added to the NH3 solution to achieve a buffer with a pH of 10.00.

Determine the amount of 0.236 M NH3 needed (Part c)

Firstly, let's find the moles of NH4+ in 2.08g of NH4Cl: moles of NH4+ = mass / molar mass moles of NH4+ = 2.08g / (14.01+1.01+35.45)g/mol ≈ 0.0284 mol Now, we use the desired ratio to find the moles of NH3 needed: moles of NH3 = moles of NH4+ / desired ratio moles of NH3 = 0.0284 mol / 5.62 ≈ 0.00505 mol Finally, using the concentration of NH3 solution given (0.236 M), we can find the required volume of NH3: volume of NH3 = moles of NH3 / concentration volume of NH3 ≈ 0.00505 mol / 0.236 M ≈ 21.4 mL Therefore, 21.4 mL of 0.236 M NH3 must be added to 2.08g of NH4Cl to achieve a pH of 10.00.

Determine the volume of 0.499 M NH3 needed (Part d)

First, calculate the moles of NH4+ in the 395 mL of 0.109 M NH4Cl solution: moles of NH4+ = volume * concentration = (395 mL)(0.109 M) ≈ 43.0 mmol Now, we use the desired ratio to find the moles of NH3 needed: moles of NH3 = moles of NH4+ / desired ratio moles of NH3 = 43.0 mmol / 5.62 ≈ 7.65 mmol Finally, with the given concentration of the NH3 solution (0.499 M), we can calculate the required volume of NH3: volume of NH3 = moles of NH3 / concentration volume of NH3 ≈ 7.65 mmol / 0.499 M ≈ 15.3 mL Thus, 15.3 mL of 0.499 M NH3 must be added to the 395 mL of 0.109 M NH4Cl solution to obtain a pH of 10.00.