Just as pH is the negative logarithm of [H3O+], p K a is the negative logarithm

Question

Just as pH is the negative logarithm of [H3O+], pKa is the negative logarithm of Ka,

pKa=logKa

The Henderson-Hasselbalch equation is used to calculate the pH of buffer solutions:
pH=pKa+log[base][acid]

Notice that the pH of a buffer has a value close to the pKa of the acid, differing only by the logarithm of the concentration ratio [base]/[acid]. The Henderson-Hasselbalch equation in terms of pOH and pKb is similar.

pOH=pKb+log[acid][base]

1.) How many grams of dry NH4Cl need to be added to 2.00 L of a 0.600 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.61? Kb for ammonia is 1.8×105.

Express your answer with the appropriate units.

Explanation / Answer

Solution:-

Using the Henderson-Hasselbalch equation:

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

Here, [A^-] = [NH3] (base)
[HA] = [NH4^+] (acid)

Kb = 1.8 x 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])
8.61 = 9.26 + log((0.600 M NH3)/(x M NH4^+))
-0.65 = log((0.600 M NH3)/(x M NH4^+))
(0.600 M NH3)/(x M NH4^+) = 10^-0.65= 0.2238
x M NH4^+ = (0.600 M NH3 )/(0.2238)
x M NH4^+ = 2.68 M NH4^+

Now needed NH4CL concentration to be 2.68 M.from question we have 2 L of solution, so the number of moles of NH4Cl we need are:

(2.68 M NH4Cl)*(2 L) = 5.36 moles of NH4Cl

(5.36 moles of NH4Cl)*(53.491 g/mol) = 286.71 g of NH4Cl

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