TEST OF PRACTICAL KNOWLEDGE QUESTION
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 HCI containing 5.0g dm\(^{-3}\). B is a solution of impure KOH containing 6.50g dm\(^{-3}\).
a) Put A into the burette and titrate it against 20.0 cm\(^3\) or 25.0 cm\(^3\) portions of B using methyl orange as indicator Tabulate your burette readings and calculate the average volume of A used. The equation for the reaction involved in the titration is: HCI\(_{(aq)}\) KOH\(_{(aq)}\) \(\to\) Cl\(_{(aq)}\) H\(_2\)O\(_{(l)}\)
(b) From your results and the information provided above, calculate the:
(i) concentration of A in mol dm\(^{-3}\)
(ii) concentration of B in mol dm\(^{-3}\)
(iii) percentage purity of KOH in B [H= 1; CI = 35.5; KOH = 56.0g mol\(^{-1}\)]
The average titre of A comes from the candidate's own burette readings. The concentration of A is fixed by the data; the concentration and purity of B follow from the titre. A representative average titre of 18.00 cm3 of A for each 25.0 cm3 portion of B is used below to complete the worked model.
(a) Tabulate rough and concordant titres and average them; here average volume of A = 18.00 cm3.
(b)(i) Concentration of A in mol dm-3
\[C_A = \frac{5.0}{36.5} = 0.137\ \text{mol dm}^{-3}\]
(ii) Concentration of B in mol dm-3
Reaction: \(HCl + KOH \rightarrow KCl + H_2O\) (mole ratio 1:1).
Amount of HCl \(= 0.137 \times \dfrac{18.00}{1000} = 2.47 \times 10^{-3}\ \text{mol}\) = amount of KOH in 25.0 cm3 of B.
\[C_B = 2.47 \times 10^{-3} \times \frac{1000}{25.0} = 0.0986\ \text{mol dm}^{-3}\]
(iii) Percentage purity of KOH in B
Mass of pure KOH in B \(= 0.0986 \times 56.0 = 5.52\ \text{g dm}^{-3}\).
B was made up as 6.50 g dm-3 of impure solid, so
\[\text{purity} = \frac{5.52}{6.50} \times 100 = 84.9\%\]
The concentration of A and the method are exact; the numerical value of B's concentration and the purity scale directly with the candidate's measured titre.
The average titre of A comes from the candidate's own burette readings. The concentration of A is fixed by the data; the concentration and purity of B follow from the titre. A representative average titre of 18.00 cm3 of A for each 25.0 cm3 portion of B is used below to complete the worked model.
(a) Tabulate rough and concordant titres and average them; here average volume of A = 18.00 cm3.
(b)(i) Concentration of A in mol dm-3
\[C_A = \frac{5.0}{36.5} = 0.137\ \text{mol dm}^{-3}\]
(ii) Concentration of B in mol dm-3
Reaction: \(HCl + KOH \rightarrow KCl + H_2O\) (mole ratio 1:1).
Amount of HCl \(= 0.137 \times \dfrac{18.00}{1000} = 2.47 \times 10^{-3}\ \text{mol}\) = amount of KOH in 25.0 cm3 of B.
\[C_B = 2.47 \times 10^{-3} \times \frac{1000}{25.0} = 0.0986\ \text{mol dm}^{-3}\]
(iii) Percentage purity of KOH in B
Mass of pure KOH in B \(= 0.0986 \times 56.0 = 5.52\ \text{g dm}^{-3}\).
B was made up as 6.50 g dm-3 of impure solid, so
\[\text{purity} = \frac{5.52}{6.50} \times 100 = 84.9\%\]
The concentration of A and the method are exact; the numerical value of B's concentration and the purity scale directly with the candidate's measured titre.
