All your burette readings (initial and final), as well as the size of your pipette, must be recorded but no account of experimental procedure is required. All calculations must be done in your answer book.
A is a solution of tetraoxosulphate (VI) acid. B is a solution containing 1.4g of potassium hydroxide per 250cm\(^3\)
(a) Put A into burette and titrate with 20.0cm\(^3\) or 25.0cm\(^3\) portions of B using methyl orange or screened methyl orange as indicator. Record the volume of your pipette. Tabulate your burette readings and calculate the average volume of A used.
(b) From your results and the information provided, calculate the:
(i) concentration of B in mol.dm\(^{-3}\)
(ii) concentration of A in mol dm\(^{-3}\)
(iii) number of hydrogen ions in 1.0dm\(^{-3}\) of A. The equation for the reaction is: H\(_2\)SO\(_4\) + 2KOH \(\to\) K\(-2\)SO\(_4\) + 2H\(_2\)O. [H= 1, O = 16; K = 39. Avogadro constant = 6.0 x 10\(^{23}\)]
(c) State whether the pH of the following would be equal to 7, greater than 7 or less than 7
(i) Solution A:
(ii) Titration mixture of A and B before the endpoint
Explanation
Volume of base used (pipette) = 25.00 cm\(^3\);
| Burette reading | Rough | 1st | 2nd | 3rd |
| Final burette readings (cm\(^3\)) | 18.60 | 18.00 | 20.30 | 28.00 |
| Initial urette readigs (cm\(^3\)) | 0.00 | 0.00 | 2.30 | 10.00 |
| Volume of acid used (cm\(^3\)) | 18.60 | 18.00 | 18.00 | 18.00 |
Average volume of the acid = \(\frac{18.00 18.00 18.00}{3}\) = 18.00 cm\(^{3}\)
(b)(i) Mass of KOH per dm\(^3\) of B = \(\frac{1.4 \times 1000}{250}\) = 5.6 g/dm\(^{-3}\)
Molar mass of KOH = 39 + 16 + 1 = 56g
Concentration of B = \(\frac{5.6}{56}\) = 0.01 mol dm\(^{-3}\)
(b)(ii) Equation for the reactions; H\(_2\)SO\(_4\) + 2KOH \(\to\) K\(_2\)SO\(_4\) + 2H\(_2\)O
From the equation,
Concentration of A (C\(_A\)) x volume of A(V\(_A\)) mole ratio \(\frac{A}{B}\)
Concentratio B(C\(_A\)) x volume of (V\(_B\))
\(\frac{C_A \times 18.00}{0.10 \times 25}\) = \(\frac{1}{2}\)
C\(_A = \frac{0.10 \times 25}{2 \times 18.00}\) = 0.0694 mol/dm\(^3\)
(b)(iii) 1 dm\(^3\) of 1.0 mol/dm\(^3\)
H\(_2\)SO\(_4\) contains (2 x 6.0 x 10\(^{23}\))
Hydrogen atom 1 dm\(^3\) of 0.0694 mol dm\(^{-3}\)
H\(_2\)SO\(_4\) contains = (2 x 6.0 x 10\(^{23}\) x 0.6694) 8.328 x 10\(^{22}\) x 10\(^{22}\)
(c)(i) Less than
(ii) Greater than; C is NaHCO\(_3\); D is Pb(NO\(_3\))\(_2\)