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SS3: CHEMISTRY - 1ST TERM

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When an impure chemical substance is used in titration, only the pure part of that substance takes part in the chemical reaction and is consumed. The impurity will not react. 

Example 1

A is 0.100 moldm-3 solution of HNO3

B is 5.00 g mixture of Na2CO3 and NaCl in 500 cm3.

Put A into a burette and titrate it against 25 cm3 portion of B using methyl orange indicator. Tabulate your burette readings and determine the average volume of A used. From your result and information provided, calculate the

i. concentration of Na2CO3 in moldm-3
ii. the concentration of Na2CO3 in gdm-3
iii. percentage purity

The equation for the reaction is: 

2HNO3(aq) + Na2CO3(aq) → 2NaNO3(aq) + H2O(l) + CO(g)

Results and Calculations

Indicator used – methyl orange

Volume of pipette – 25 cm3

Burette reading (cm3)Rough1st2nd
Final burette reading (cm3)33.7033.4033.50
Initial burette reading (cm3)0.000.000.00
Volume of acid used (cm3)33.7033.4033.50

The average titre value = \( \frac{3.30 \; + \; 33.50}{2} \scriptsize cm^3 \\ = \scriptsize 33.45cm^3 \)

(i). The mole of acid used = \( \frac{C \;\times \; V}{100} \\ = \frac{0.100 \;\times \; 33.45}{100} \\ = \scriptsize 0.00335 mol\)

From the equation of the reaction;

2 mol of HNO3 requires 1mol of Na2CO3

Hence, 0.00335 mol of HNO3 will require

\( \frac{1\;\times \; 0.00335}{2} \)

= 0.00168 mol of Na2CO3

25 cm3 of B contains 0.00168 mol of Na2CO3

1000 cm3 of B will contain

\( \frac{0.00168 \;\times \; 1000}{25} \)

 = 0.0672 mol dm-3

Alternatively,

\( \frac{C_A V_A}{C_B V_B} = \frac{n_A}{n_B}\)

Substituting the values, we get:

\( \frac{0.1 \; \times \; 33.45}{C_B \; \times \; 25} = \frac{2}{1}\\ \scriptsize C_B = \normalsize \frac{0.1 \; \times \; 33.45 \; \times \; 1}{25 \; \times \; 2} \\ = \scriptsize 0.00672moldm^{-3}\)

(ii). The molar concentration of Na2CO3 in the solution of Bis 0.0672 moldm-3

Molar mass of Na2CO3 = 106gmol-1

∴Mass concentration of Na2CO3 = 0.0672 moldm-3 x 106 gmol-1 = 7.12g dm-3.

(iii). 500 cm3 of B contains 5.00g of the mixture

1000 cm3 will contain

\( \frac{1000 \; \times \; 5}{500}\\ \scriptsize = 10.0gdm^{-3} \)

Therefore %purity =

\( \frac{concentration \; of \; pure}{concentration \; of \; impure} \; \times \; \frac{100}{1}\)

= \( \frac{7.12}{10} \; \times \; \frac{100}{1} \\ = \scriptsize 71.2 \%\)

Now, %impurity = 100 – %purity

= 100 – 71.2 = 28.8%

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