Question

Calculate the concentration of HTart anion in this solution
0.500M is for KNO3

KNO3(aq) + KHTart(s) -> KNO3(aq) + K(aq) + HTart (aq)

Originally, Htart was KHTart (solid) and had 2.0 added into 150mL of the KNO3(aq) solution.
How do you get the concentration of HTart?

Answer 1

Into 150 mL of the 0.50 M KNO₃ (aq) solution was added 2.0 g of KHTart (solid).

Calculate the concentration of HTart⁻ anion in this solution.

1. Calculating mass of KNO₃ solution:

$m_{sol} = V \cdot \rho$, where $V$ – volume of solution, $\rho = 1.029\ \mathrm{g/mL}$ – density of 0.5 M KNO₃ (aq) solution (from the chemical handbook); $m_{sol} = 150 \cdot 1.029 = 154.35\ \mathrm{g}$;

2. Calculating mass of solid KNO₃, dissolved in water:

$m_{KNO3} = V \cdot C \cdot M_{KNO3} / 1000$, where $C$ – molar concentration of KNO₃ in solution, $M_{KNO3} = 101\ \mathrm{g/mol}$;

$m_{KNO3} = 150 \cdot 0.5 \cdot 101 / 1000 = 7.58\ \mathrm{g}$;

3. Calculating mass of pure water:

$m_{H_2O} = m_{sol} - m_{KNO3} = 154.35 - 7.58 = 146.77\ \mathrm{g}$;

4. Calculating mass of KHTart, which will dissolve in such amount of water:

$m_{KHTart} = m_{H_2O} \cdot S_{KHTart} / 100$, where $S_{KHTart} = 0.54\ \mathrm{g}$ – solubility of KHTart in 100 g of water at 20°C (from the chemical handbook); $m_{KHTart} = 146.77 \cdot 0.54 / 100 = 0.793\ \mathrm{g}$

5. Calculating the molar concentration of HTart⁻ anion in this solution, it will be the same as the molar concentration of dissolved salt KHTart, because that is a strong electrolyte, which fully dissociates:

$C_{HTart^{-}} = C_{KHTart} = \frac{m_{KHTart} \cdot 1000}{V \cdot M_{KHTart}} = \frac{0.793 \cdot 1000}{150 \cdot 188} = 0.0281\ \mathrm{M}$

Answer: 0.0281 mol/L