How many grams of ammonium chloride, NH4Cl, would have to be added to exactly?

Question

500 mL of 0.10 M NH3 solution to have a pH of 9.0?

Answer 1

We know the solution is a buffer solution, because NH3 is a base and NH4+ is its conjugate acid. We can use the Henderson Hasselbach equation to find out the concentration of of NH4+ (acid) for a solution with a pH of 9.

Ka of NH4+ is 5.6*10^-10 (I found this out by looking it up)

[base] = concentration of base (NH3)

[acid] = concentration of acid (NH4+)

pH = pKa + log([base]/[acid])

9 = -log(5.6*10^-10) + log([NH3]/[NH4+])

9 = 9.2518 + log(.1M/[NH4+])

-.2518 = log(.1M/[NH4+])

10^-.2518 = .1M/[NH4+]

.56 = .1M/[acid]

[NH4+] = .1/.56

[NH4+] = .1786M

Now we know the concentration of NH4+ need in the buffer solution. Now we just use stoichiometry. For every one mole of NH4+ in solution there is one mole of NH4Cl needed.

.5L of NH4+ buffer *(.1786moles/L) *(1mole NH4Cl/1mole NH4+) *(53.5g/mol) = 4.78g of NH4Cl needed