# Question a4463

Jul 2, 2017

You should add 4,5 g of $\text{NH"_4"Cl}$.

#### Explanation:

$\text{NH"_3 + "H"_2"O" ⇌ "NH"_4^"+" + "OH"^"-}$

Write the equilibrium constant expression

${K}_{\textrm{b}} = \left(\left[{\text{NH"_4^"+"]["OH"^"-"])/(["NH}}_{3}\right]\right)$

Solve for the concentration of hydroxide ion

["OH"^"-"] = K_text(b) × (["NH"_3])/(["NH"_4^"+"])

Notice that this expression involves the ratio of the concentrations of the base and its conjugate acid.

However, the concentrations are moles divided by litres, and both species are in the same solution, so they are in the same number of litres.

Hence, the ratio of the molarities is the same as the ratio of the moles.

If both species are in the same solution, you can use either moles or molar concentrations, whichever is most convenient.

Now, let's solve the problem.

If $\text{pH = 9, pOH = 14.00 - 9 = 5}$

Then ["OH"^"-"] = 10^"-5"color(white)(l)"mol/L"

"1,8" × color(red)(cancel(color(black)(10^"-5"))) = color(red)(cancel(color(black)(10^"-5"))) × "0,5 mol/L"/(["NH"_4^"+"])

["NH"_4^"+"] = "0,5 mol/L"/1.8 = "0,28 mol/dm"^3#

The $\text{NH"_4^"+}$ is in the same solution as the ${\text{NH"_3: "0,300 dm}}^{3}$.

$\text{Moles of NH"_4^"+" = "0,300" color(red)(cancel(color(black)("dm"^3))) × "0,28 mol"/(1 color(red)(cancel(color(black)("dm"^3)))) = "0,083 mol}$

$\text{Moles of NH"_4"Cl" = "Moles of NH"_4^"+" = "0,083 mol}$

$\text{Mass of NH"_4"Cl" = "0,083" color(red)(cancel(color(black)("mol"))) × "53,49 g"/(1 color(red)(cancel(color(black)("mol")))) = "4,5 g}$