Answer:
a. pH = 9.26
b. pH = 8.87
c. pH = 1.76
Explanation:
a. Using Henderson-Hasselbalch formula:
pH = pka + log [A⁻] / [HA]
Where A⁻ is NH₃ and HA is NH₄Cl
The ka of NH₃/NH₄⁺ buffer is kw/kb = 1x10⁻¹⁴ / 1,8x10⁻⁵
ka = 5,56x10⁻¹⁰
pka = -log ka = 9.26
pH = pka + log [0,10] / [0,10]
<em>pH = 9.26</em>
b. Reaction of NH₃ with HCl is:
NH₃ + HCl → NH₄Cl
Initial moles of NH₃ and NH₄Cl are:
0,125L×(0.10mol / 1L) = 0.0125 moles
Moles of HCl are:
0.012L×(0.20mol / 1L) = 0.0024 moles
That means moles produced of NH₄Cl are 0.0024 moles and total moles are:
0.0125 moles + 0.0024 moles = 0.0149 moles of NH₄Cl
And moles of NH₃ are:
0.0125 moles - 0.0024 moles = 0.0101 moles of NH₃
pH is:
pH = 9.26 + log [0.0101] / [0.0149]
<em>pH = 8.87</em>
c. The moles of H⁺ are:
0.012L×(0.20mol / L) = 2.4x10⁻³ moles
Total volume is 125mL + 12mL = 137mL
[H⁺] = 2.4x10⁻³ moles / 0.137L = 0.0175M
pH = -log [H⁺] = <em>1.76</em>
If you compare the change in pH of b and a with pH in c you can see the effect of a buffer that regulate the pH in a solution.
I hope it helps!
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