a) (i) What is meant by the term buffer solution? A buffer solution is a solution that resists or minimizes a change in its pH when small amounts of acid or base are added - CIE - A-Level Chemistry - Question 4 - 2015 - Paper 4

The control of blood pH by the hydrogencarbonate ion can be illustrated by the following equilibria:

1. $HCO_3^- + H^+ \leftrightarrow H_2CO_3$
2. $H_2CO_3 \leftrightarrow H_2O + CO_2$

These reactions illustrate how HCO₃⁻ can accept a proton (H⁺) to form carbonic acid (H₂CO₃), thus regulating acidity.

To calculate the pH of the buffer solution, we will use the Henderson-Hasselbalch equation:

$pH = pK_a + \log\left(\frac{[base]}{[acid]}\right)$

First, calculate the pKₐ for the equilibrium given:

$pK_a = -\log(6.2 \times 10^{-3}) = 2.21$

Next, determine the concentrations of the base (Na₂HPO₄) and the acid (NaH₂PO₄) in the mixed solution. Since the volumes are equal, we can average the concentrations:

Total volume = 200 cm³ = 0.2 dm³

The concentration of Na₂HPO₄:

$[base] = \frac{0.5 \text{ mol/dm}^3 \times 100 \text{ cm}^3}{200 \text{ cm}^3} = 0.25 \text{ mol/dm}^3$

The concentration of NaH₂PO₄:

$[acid] = \frac{0.3 \text{ mol/dm}^3 \times 100 \text{ cm}^3}{200 \text{ cm}^3} = 0.15 \text{ mol/dm}^3$

Now substituting back into the Henderson-Hasselbalch equation:

$pH = 2.21 + \log\left(\frac{0.25}{0.15}\right) = 2.21 + 0.125$

Thus, the pH = 2.34.

The solubility product expression for silver phosphate, Ag₃PO₄, is given by:

$K_{sp} = [Ag^+]^3[PO_4^{3-}]$

The units of Ksp are mol⁴ dm⁻¹. This is derived from the concentration terms raised to the power of the coefficients in the balanced equation.

Given that the Ksp value is 1.25 x 10⁻² at 298K, and knowing that, in a saturated solution:

$[Ag^+] = 3x$ $[PO_4^{3-}] = x$

We can substitute:

$K_{sp} = (3x)^3(x) = 27x^4$

Setting this equal to the Ksp value gives:

$27x^4 = 1.25 \times 10^{-2}$

Solving for x:

$x^4 = \frac{1.25 \times 10^{-2}}{27} = 4.63 \times 10^{-4}$

Thus, taking the fourth root:

$x = (4.63 \times 10^{-4})^{1/4} = 3.39 \times 10^{-2} \text{ mol dm}^{-3}$

So, $[Ag^+] = 3x = 1.02 \text{ mol dm}^{-3}$.

The half-equation for the reaction can be derived from the redox properties of iron and phosphoric acids. If we denote the reduction reaction:

$Fe^{3+} + e^- \rightarrow Fe^{2+}$

The overall reaction can be expressed as:

$H₃PO₃ + 2Fe^{3+} + 2e^- \rightarrow H₃PO₄ + 2Fe^{2+}$

The standard cell potential can be calculated using:

$E°_{cell} = E°_{reduction} - E°_{oxidation}$ For our case:

$E°_{cell} = -0.28 - (0.77) = ...$

Thus calculating E° and including the values gives the final answer.