Question

# What mass of ammonium chloride must be added to exactly 500 ml of .10 m NH3...

What mass of ammonium chloride must be added to exactly 500 ml of .10 m NH3 solution to give a solution with a pH of 9.00?

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 5.6x10-10
pKa = -log(5.6x10-10) = 9.25

[base] = concentration of base (NH3)
[acid] = concentration of acid (NH4+)

pH = pKa + log([base]/[acid])
9 = 9.25 + log([NH3]/[NH4+])
9 = 9.25 + log(0.1/[NH4+])
-0.25 = log(0.1/[NH4+])
10(-0.25) = 0.1/[NH4+]
.56 = 0.1M/[acid]
[NH4+] = 0.1786 M

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 *(0.1786moles/L) *(1mole NH4Cl/1mole NH4+) *(53.5g/mol) = 4.78g of NH4Cl needed

Hope this helps

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