Respuesta :
Answer:
The correct answer is 574.59 grams.
Explanation:
Based on the given information, the number of moles of NH₃ will be,
= 2.50 L × 0.800 mol/L
= 2 mol
The given pH of a buffer is 8.53
pH + pOH = 14.00
pOH = 14.00 - pH
pOH = 14.00 - 8.53
pOH = 5.47
The Kb of ammonia given is 1.8 * 10^-5. Now pKb = -logKb,
= -log (1.8 ×10⁻⁵)
= 5.00 - log 1.8
= 5.00 - 0.26
= 4.74
Based on Henderson equation:
pOH = pKb + log ([salt]/[base])
pOH = pKb + [NH₄⁺]/[NH₃]
5.47 = 4.74 + log ([NH₄⁺]/[NH₃])
log([NH₄⁺]/[NH₃]) = 5.47-4.74 = 0.73
[NH₄⁺]/[NH₃] = 10^0.73= 5.37
[NH₄⁺ = 5.37 × 2 mol = 10.74 mol
Now the mass of dry ammonium chloride required is,
mass of NH₄Cl = 10.74 mol × 53.5 g/mol
= 574.59 grams.
The amount of dry NH4Cl in grams is 568.28 grams
The Henderson Hasselbalch equation is an approximation that demonstrates the relation between a solution's pH or pOH, its pKa or pKb, and the concentration ratio of the dispersed chemical species.
In other words, the Henderson-Hasselbalch equation is also utilized when calculating the pH equilibrium of an acid-base reaction as well as the pH of a buffer solution.
It is usually expressed by using the relation:
[tex]\mathbf{pOH = pKb + log \dfrac{[salt]}{[base]}}[/tex]
[tex]\mathbf{14 - pH =- log Kb + log \dfrac{[NH_4Cl]}{[NH_3]}}[/tex]
replacing the values from the given information, we have:
[tex]\mathbf{14 - 8.53 =- log (1.8\times 10^{-5}) + log \dfrac{[NH_4Cl]}{[NH_3]}}[/tex]
[tex]\mathbf{5.47=4.7447 + log \dfrac{[NH_4Cl]}{[NH_3]}}[/tex]
[tex]\mathbf{log \dfrac{[NH_4Cl]}{[NH_3]} = 5.47-4.7447 }[/tex]
[tex]\mathbf{ \dfrac{[NH_4Cl]}{[NH_3]} = log^{-1}(0.7253 )}[/tex]
[tex]\mathbf{ \dfrac{[NH_4Cl]}{[NH_3]} = 5.312 }[/tex]
By cross multiply;
[tex]\mathbf{moles \ of \ NH_4Cl = moles \ of \ NH_3 \times 5.312 }[/tex]
where;
- moles of NH₃ = molarity × volume
- moles of NH₃ = 0.8 × 2.50
- moles of NH₃ = 2
∴
[tex]\mathbf{moles \ of \ NH_4Cl = 2 \ moles \times 5.312 }[/tex]
[tex]\mathbf{moles \ of \ NH_4Cl =10.624 \ moles }[/tex]
Recall that:
The number of moles = mass/molar mass.
∴
- The mass of NH₄Cl = number of moles × molar mass
- The mass of NH₄Cl = 10.624 moles × 53.491 g/mol
- The mass of NH₄Cl = 568.28 grams
Therefore, we can conclude that the amount of dry NH4Cl in grams is 568.28 grams
Learn more about Henderson Hasselbalch equation here:
https://brainly.com/question/9129423?referrer=searchResults
