TEST OF PRACTICAL KNOWLEDGE QUESTION All yo..

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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 the experimental procedure is required. All calculations must be done in your answer booklet.

A is a solution containing 5.00 g of HNO$_3$ in 500 cm\(^3) of solution. B is a solution of NaOH of unknown concentration.

(a) Put A into the burette and titrate it with 20.0 cm$^3$ or 25.0 cm$^3$ portions of B using methyl orange as an indicator. Repeat the titration to obtain concordant titre values. Tabulate your results and calculate the average volume of acid used. Equation of the reaction is HNO$_{3(aq)}$ + NaOH$_{(aq)}$ $\to$ NaNO$_{3(aq)}$ + H$_2$O$_{(l)}$

(b) From your results and the information provided. calculate the: (i) concentration ot A In mol dm$^{-3}$

(ii) concentration of B in mol dm$^{-3}$.

(iii) concentration of B in gdm$^{-3}$

(iv) mass of NaNO$_3$ formed. If 250 cm$^3$ of NaOH were neutralised. [Molar mass of NaOH = 40g mol$^{-1}$, NaNO$_3$ = 85 gmol$^{-1}$. Credit will be given for strict adherence to the instructions. for observations precisely recorded and for accurate inferences. All tests, observations and inferences must be clearly entered in this booklet, in ink, at the time they are made.

Answer

Indicator = Methyl Orange

Volume of the base used = 25.00cm $^{3}$

Titration	Rough Titre	1st Titre	2nd Titre	3rd Titre
Final burette readings cm $^{3}$	24.70	24.80	24.70	24.90
Initial burette reading cm $^{3}$	0.00	0.00	0.00	0.00
Volume of acid used cm $^{3}$	24.70	24.80	24.70	24.90

$\frac{24.80 + 24.70 + 24.90}{3}$

Average Titre = $\frac{1 s t + 2 n d + 3 r d}{3}$

= $\frac{24.80 + 24.70 + 24.90}{3}$

$\frac{24.80 + 24.70 + 24.90}{3}$

= 24.80cm $^{3}$

Alternatively 2 concordant titres can be used to calculate average titre

Equation of thereaction; Na $_{2}$ CO $_{3}$ .XH $_{2}$ O + 2HCl $_{(a q)}$ $\to$ 2Nacl $_{(a q)}$ $\to$ 2Nacl $_{(a q)}$ + CO $_{2 (a q)}$ + (X + 1)H $_{2}$ O $_{(l)}$

$\frac{C_{A} V_{A}}{C_{B} V_{B}} = \frac{n_{A}}{n_{B}}$

C $_{A}$ = Molar concentration of HCl $_{(a q)}$ in moldm $^{3}$

V $_{A}$ = Volume of acid used in cm $^{3}$ = 24.80cm $^{3}$

C $_{B}$ = Molar concentration of Na $_{2}$ Cu $_{3}$ . xH $_{2}$ O in molddm $^{3}$

n $_{A}$ = 2

n $_{B}$ = 1

(b)(i) Concentration of C in moldm $^{- 3}$ = ?

From the equation reaction;

$\frac{C_{A} V_{A}}{C_{B} V_{B}} = \frac{n_{A}}{n_{B}}$

C $_{A}$ = 0.200 molddm $^{- 3}$

C $_{B}$ = ?

V $_{A}$ = 24.80 cm

V $_{B}$ = 25.00 cm $^{3}$

Substitution of known values

$\frac{0.200 \times 24.80}{C_{B} \times 25.00} = \frac{2}{1}$

C $_{B}$ = $\frac{1 \times 0.200 \times 24.80}{2 \times 25.00}$

C $_{B}$ = 0.0992 moldm $^{- 3}$

(ii) Concentration of C in gdm $^{3}$ = ?

500 cm $^{2}$ $\to$ 14.3 g

1000cm $^{3}$ $\to$ $\frac{14.3}{500}$ x 1000g

= 28.6 dm $^{- 3}$

(iii) Molar mass of Na $_{2}$ CO $_{3}$ . XH $_{2}$ O

Molar conc in moldm $^{- 3}$ = $\frac{{conc. in gdm}^{- 3}}{molar mass}$

Molar mass gmol $^{- 1}$ = $\frac{{conc in g dm}^{- 3}}{{molar conc.in moldm}^{- 3}}$

$\frac{28.6 g d m^{- 3}}{0.0992 m o l d m^{- 3}}$

= 288.3065

= 288 gmol $^{- 1}$

(iv) Value of x in Na $_{2}$ CO $_{3}$ . xH $_{2}$ O?

[H = 1.0, C = 12.0, O = 16.0, Na = 23.0]

N $_{2}$ CO $_{3}$ . xH $_{2}$ O = 288

2(23) + 12 + 3 (16) + x(2(1) + 16) = 288

46 + 12 +48 +18x = 288

106 + 18x = 288

18x = 288 - 106

18x = 182

x = $\frac{182}{18}$

x = 10.11

x = 10

Read lesson note on Stoichiometry And Chemical Reactions (WAEC)

Answer Details

Indicator = Methyl Orange

Volume of the base used = 25.00cm $^{3}$

Titration	Rough Titre	1st Titre	2nd Titre	3rd Titre
Final burette readings cm $^{3}$	24.70	24.80	24.70	24.90
Initial burette reading cm $^{3}$	0.00	0.00	0.00	0.00
Volume of acid used cm $^{3}$	24.70	24.80	24.70	24.90

$\frac{24.80 + 24.70 + 24.90}{3}$

Average Titre = $\frac{1 s t + 2 n d + 3 r d}{3}$

= $\frac{24.80 + 24.70 + 24.90}{3}$

$\frac{24.80 + 24.70 + 24.90}{3}$

= 24.80cm $^{3}$

Alternatively 2 concordant titres can be used to calculate average titre

Equation of thereaction; Na $_{2}$ CO $_{3}$ .XH $_{2}$ O + 2HCl $_{(a q)}$ $\to$ 2Nacl $_{(a q)}$ $\to$ 2Nacl $_{(a q)}$ + CO $_{2 (a q)}$ + (X + 1)H $_{2}$ O $_{(l)}$

$\frac{C_{A} V_{A}}{C_{B} V_{B}} = \frac{n_{A}}{n_{B}}$

C $_{A}$ = Molar concentration of HCl $_{(a q)}$ in moldm $^{3}$

V $_{A}$ = Volume of acid used in cm $^{3}$ = 24.80cm $^{3}$

C $_{B}$ = Molar concentration of Na $_{2}$ Cu $_{3}$ . xH $_{2}$ O in molddm $^{3}$

n $_{A}$ = 2

n $_{B}$ = 1

(b)(i) Concentration of C in moldm $^{- 3}$ = ?

From the equation reaction;

$\frac{C_{A} V_{A}}{C_{B} V_{B}} = \frac{n_{A}}{n_{B}}$

C $_{A}$ = 0.200 molddm $^{- 3}$

C $_{B}$ = ?

V $_{A}$ = 24.80 cm

V $_{B}$ = 25.00 cm $^{3}$

Substitution of known values

$\frac{0.200 \times 24.80}{C_{B} \times 25.00} = \frac{2}{1}$

C $_{B}$ = $\frac{1 \times 0.200 \times 24.80}{2 \times 25.00}$

C $_{B}$ = 0.0992 moldm $^{- 3}$

(ii) Concentration of C in gdm $^{3}$ = ?

500 cm $^{2}$ $\to$ 14.3 g

1000cm $^{3}$ $\to$ $\frac{14.3}{500}$ x 1000g

= 28.6 dm $^{- 3}$

(iii) Molar mass of Na $_{2}$ CO $_{3}$ . XH $_{2}$ O

Molar conc in moldm $^{- 3}$ = $\frac{{conc. in gdm}^{- 3}}{molar mass}$

Molar mass gmol $^{- 1}$ = $\frac{{conc in g dm}^{- 3}}{{molar conc.in moldm}^{- 3}}$

$\frac{28.6 g d m^{- 3}}{0.0992 m o l d m^{- 3}}$

= 288.3065

= 288 gmol $^{- 1}$

(iv) Value of x in Na $_{2}$ CO $_{3}$ . xH $_{2}$ O?

[H = 1.0, C = 12.0, O = 16.0, Na = 23.0]

N $_{2}$ CO $_{3}$ . xH $_{2}$ O = 288

2(23) + 12 + 3 (16) + x(2(1) + 16) = 288

46 + 12 +48 +18x = 288

106 + 18x = 288

18x = 288 - 106

18x = 182

x = $\frac{182}{18}$

x = 10.11

x = 10

Read lesson note on Stoichiometry And Chemical Reactions (WAEC)

Stoichiometry And Chemical Reactions

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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 ...