Answers:
25. See below
26. 9.02
27. 0.09 mol·L⁻¹
Step-by-step explanation:
25. Buffer Solutions
a. <em>NH₄Br + NH₃
</em>
Yes. A solution of a weak base (NH₃) and its conjugate acid (NH₄⁺) is a buffer.
b. <em>Excess NaOH + HCl
</em>
<em>No</em>. A solution of a strong base is not a buffer.
c. <em>CH₃COOH + HCl
</em>
No. The strong acid will overpower the weak acid. A solution of a strong acid is not a buffer
d. <em>Excess HCl + NH₃</em>
<em>No</em>. The strong acid will overpower the weak acid, NH₄⁺.
26. pH of buffer
<em>Data:
</em>
V = 1 L
[NH₃] = 0.1 mol·L⁻¹
Mass of NH₄Cl = 96.3 g
pKₐ = 9.25
Calculations:
(a) <em>Moles of NH₄Cl
</em>
n = 96.3 g × (1 mol/56.49 g)
= 1.705 mol
(b) [NH₄⁺]
[NH₄⁺] = 1.705 mol/1 L
= 1.705 mol·L⁻¹
(c). <em>Chemical equation
</em>
NH₃ + H₂O ⇌ NH₄⁺ + OH⁻
(base) (conj. acid)
(d) <em>pH of buffer
</em>
The <em>Henderson-Hasselbalch equation</em> is
pH = pKₐ + log([base]/[acid])
pH = 9.25 + log[(0.1)/0.1705]
= 9.25 + log0.59
= 9.25 – 0.23
= 9.02
27. [NH₄⁺] in buffer
<em>Data:
</em>
[NH₃] = 0.5 mol·L⁻¹
pH = 10
<em>Calculation:
</em>
pH = pKₐ + log([base]/[acid])
10 = 9.25 + log(0.5/[NH₄⁺]) Subtract 9.25 from each side
0.75 = log(0.5/[NH₄⁺]) Take the antilog of each side
10^(0.75) = 0.5/[NH₄⁺]
5.62 = 0.5/[NH₄⁺] Multiply each side by [NH₄⁺]
5.62[NH₄⁺] = 0.5 Divide each side by 5.62
[NH₄⁺] = 0.09 mol·L⁻¹
