How many grams of dry NH4Cl need to be added to 2.00 L of a 0.800 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.63? Kb for ammonia is 1.8×10−5.

Mass NH4Cl = 365 grams

Explanation:

Step 1: Data given

Volume = 2.00 L

Molarity = 0.800M

pH = 8.63

Kb ammonia is 1.8*10^-5

Step 2: Calculate concentration NH4+

pH = pKa + log([A^-]/[HA])

Here, [A^-] = [NH3] (base)

[HA] = [NH4^+] (acid)

Rather than use Kb, it is easier to use the pKa of NH4^+ (the conjugate acid).

Kb = 1.8 * 10^-5  

pKb = -log(Kb) = -log(1.8 x 10^-5) = 4.74

pKa + pKb = 14

pKa = 14 - pKb = 14 - 4.74 = 9.26

pH = pKa + log([A^-]/[HA])

8.63 = 9.26 + log((0.800 M NH3)/(x M NH4^+))

-0.63 = log((0.800 M NH3)/(x M NH4^+))

(0.800 M NH3)/(x M NH4^+) = 10^-0.63 = 0.2344

x M NH4^+ = (0.800 M NH3 )/(0.2344)

x M NH4^+ = 3.413 M NH4^+

Step 3: Calculate moles NH4Cl

Moles NH4Cl = molarity * volumes

Moles NH4Cl = 3.413 M * 2.00 L

Moles NH4Cl = 6.826 moles NH4Cl

Step : Calculate mass NH4Cl

Mass NH4Cl = moles NH4Cl * molar mass NH4Cl

Mass NH4Cl = 6.826 moles * 53.49 g/mol

Mass NH4Cl = 365 grams

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