All your burette readings (initial and final), as well as the size of your pipette, must be recorded but on no account of experimental procedure is required. All calculations must be done in your answer book.
A is solution of trioxonitrate (V) acid, B is a solution containing 6.90 g of potassium trioxocarbonate (IV) per dm\(^3\)
(a) Put A into the buret and titrate it against 20.0 cm\(^3\) or 25.0 cm\(^3\) portions of B using methy orange or screened methyl orange as indicater. Record the volume of your pipette. Tabulate your burette readings and calculate the average volume A used. The equation for the reaction is \(\mathrm{K_2CO_{3(aq)} + 2HNO_{3(aq)} \to 2KNO_{3(aq)} + CO_{2(g)} + H_2O_{(l)}}\)
(b) From your results and the information provided calculate;
(i) concenytration of solution B in mol dm\(^{-3}\)
(ii) number of potassium ions in 1.00 dm\(^3\) of B [C = 12.0, O = 16.0, K = 39.0, Avogadro constant = 6.02 x 10\(^{23}\) mol \(^{-1}\)]
(a) The actual burette readings and average titre of A are obtained in the laboratory and are not needed for the calculations in (b), which depend only on the stated concentration of B.
(b)(i) Concentration of B in mol dm-3
Molar mass of K2CO3:
\[ M = 2(39.0) + 12.0 + 3(16.0) = 78.0 + 12.0 + 48.0 = 138.0\ \text{g mol}^{-1} \]
Concentration:
\[ C = \frac{\text{mass per dm}^3}{M} = \frac{6.90}{138.0} = 0.0500\ \text{mol dm}^{-3} \]
(b)(ii) Number of potassium ions in 1.00 dm3 of B
Each formula unit of K2CO3 releases 2 K+ ions, so:
\[ n(\text{K}^+) = 2 \times 0.0500 = 0.100\ \text{mol} \]
\[ N(\text{K}^+) = 0.100 \times 6.02\times10^{23} = 6.02\times10^{22}\ \text{ions} \]
Answer: concentration of B = 0.0500 mol dm-3; number of K+ ions in 1.00 dm3 = 6.02 × 1022.
(a) The actual burette readings and average titre of A are obtained in the laboratory and are not needed for the calculations in (b), which depend only on the stated concentration of B.
(b)(i) Concentration of B in mol dm-3
Molar mass of K2CO3:
\[ M = 2(39.0) + 12.0 + 3(16.0) = 78.0 + 12.0 + 48.0 = 138.0\ \text{g mol}^{-1} \]
Concentration:
\[ C = \frac{\text{mass per dm}^3}{M} = \frac{6.90}{138.0} = 0.0500\ \text{mol dm}^{-3} \]
(b)(ii) Number of potassium ions in 1.00 dm3 of B
Each formula unit of K2CO3 releases 2 K+ ions, so:
\[ n(\text{K}^+) = 2 \times 0.0500 = 0.100\ \text{mol} \]
\[ N(\text{K}^+) = 0.100 \times 6.02\times10^{23} = 6.02\times10^{22}\ \text{ions} \]
Answer: concentration of B = 0.0500 mol dm-3; number of K+ ions in 1.00 dm3 = 6.02 × 1022.
