This question is about sodium and some of its compounds - AQA - A-Level Chemistry - Question 8 - 2018 - Paper 1

The reaction of sodium with water can be represented as:

$2 ext{Na} + 2 ext{H}_2 ext{O} \rightarrow 2 ext{NaOH} + ext{H}_2$

To calculate the volume of hydrogen gas produced, we use the Ideal Gas Law:

$PV = nRT$

Where:

• P = 101 kPa = 101,000 Pa
• R = 8.31 J K⁻¹ mol⁻¹
• T = 298 K (25 °C)
• n = moles of hydrogen gas produced (0.0109 moles)

Thus, we rearrange to calculate volume:

$V = \frac{nRT}{P} = \frac{(0.0109 \, \text{mol})(8.31 \, \text{J K}^{-1} \text{mol}^{-1})(298 \, ext{K})}{101000 \, ext{Pa}} = 0.267 \, \text{m}^3 = 267 \, \text{cm}^3$

The concentration of sodium ions can be found using the following calculation:

$\text{Concentration} = \frac{n}{V}$

Where:

• n = 0.0109 moles of sodium ions
• V = total volume of solution (500 cm³ = 0.5 dm³)

Thus:

$\text{Concentration} = \frac{0.0109}{0.5} = 0.0218 \, \text{mol dm}^{-3}$

The shape of the NH₂⁻ ion is based on the VSEPR theory. It has a trigonal pyramidal shape due to the lone pair of electrons on the nitrogen atom influencing the bond angles.

The bond angle for NH₂⁻ is predicted to be approximately 107 degrees. This is due to the presence of the lone pair, which exerts a greater repulsive force compared to bonding pairs, slightly reducing the bond angle from the ideal tetrahedral angle of 109.5 degrees.