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]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D)
<em>1.</em> ka of 1,8x10⁻⁵ ≡ 4,74
pH = 4,74 + log₁₀
= <em>3,74</em>
pH = 4,74 + log₁₀
= <em>4,74</em>
pH = 4,74 + log₁₀
= <em>5,74</em>
<em>2. </em>Using:
pOH = pkb + log₁₀![\frac{[acid]}{[base]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Bacid%5D%7D%7B%5Bbase%5D%7D)
A pH of 8,89 is a pOH of 14-8,89 = 5,11
Thus:
5,11 = 4,74 + log₁₀![\frac{[acid]}{[base]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Bacid%5D%7D%7B%5Bbase%5D%7D)
2,32 = ![\frac{[acid]}{[base]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Bacid%5D%7D%7B%5Bbase%5D%7D)
The moles of ammonia (base) are:
2,20L × 0,400M = 0,88 moles
Replacing:
2,32 = ![\frac{[acid]}{[0,88]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Bacid%5D%7D%7B%5B0%2C88%5D%7D)
[acid] = 2,0416 moles of NH₄Cl ₓ (53,491g/mol) = <em>109,2 g of NH₄Cl</em>
You need to add 109,2g of NH₄Cl.
I hope it helps!