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] Part APart complete Acetic acid has a Ka of 1.8×10−5. Three acetic acid/acetate buffer solutions, A, B, and C, were made using varying concentrations: [acetic acid] ten times greater than [acetate], [acetate] ten times greater than [acetic acid], and [acetate]=[acetic acid]. Match each buffer to the expected pH. Drag the appropriate items to their respective bins. View Available Hint(s) ResetHelp pH = 3.74 [acetic acid] ten times greater than [acetate] pH = 4.74 [acetate] = [acetic acid] pH = 5.74 [acetate] ten times greater than [acetic acid] Previous Answers Correct Part B How many grams of dry NH4Cl need to be added to 2.20 L of a 0.400 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.89? Kb for ammonia is 1.8×10−5

Answer 1

Answer:

You need to add 109,2g of NH₄Cl.

Explanation:

To calculate the pH in a buffer you can use Henderson-Hasselbalch formula:

pH = pka + log₁₀$\frac{[base]}{[acid]}$

1. ka of 1,8x10⁻⁵ ≡ 4,74

pH = 4,74 +  log₁₀$\frac{[1]}{[10]}$ = 3,74

pH = 4,74 +  log₁₀$\frac{[1]}{[1]}$ = 4,74

pH = 4,74 +  log₁₀$\frac{[10]}{[1]}$ = 5,74

2. Using:

pOH = pkb + log₁₀$\frac{[acid]}{[base]}$

A pH of 8,89 is a pOH of 14-8,89 = 5,11

Thus:

5,11 = 4,74 +  log₁₀$\frac{[acid]}{[base]}$

2,32 = $\frac{[acid]}{[base]}$

The moles of ammonia (base) are:

2,20L × 0,400M = 0,88 moles

Replacing:

2,32 = $\frac{[acid]}{[0,88]}$

[acid] = 2,0416 moles of NH₄Cl ₓ (53,491g/mol) = 109,2 g of NH₄Cl

You need to add 109,2g of NH₄Cl.

I hope it helps!