Quantitative Analysis II
Calculations

Example 1: In a titration experiment, 21.60 cm³ of H2SO4(aq)was required to completely react with 25.0 cm³ of NaOH(aq) containing 2.00 g of the alkali in 500 cm³ of solution.

  1. Write a balanced equation for the reaction.
  2. Calculate the concentration of:
    1. NaOH(aq) in moldm-3
    2. H2SO4(aq) in moldm-3
    3. H2SO4(aq) in gdm-3
  3. State three precautions that should be taken during the titration. [H = 1.0, O = 16.0, Na = 23.0, S = 32.0] [ WAEC ]

Solution

i. The balanced equation can be given as

$${H_2SO_4} + {2NaOH} ——> {Na_2SO_4} + {2H_2O} $$

b (i) Concentration of NaOH in mol/dm³ $$\text{Conc in mol/dm³} = \frac{\text {Conc. in g/dm³}}{\text{molar mass}} $$
Concentration of NaOH given = 2g per 500cm³

Convert this concentration to g/dm³
$$\text{if 2g of NaOH} ———> {500cm³}$$ $$\text{Xg of NaOH} ——> {1000cm³}{(1dm³)}$$ $$\text{Conc. of NaOH in g/dm³} = \frac{2 × 1000}{500} $$ $$\text{Conc. of NaOH in g/dm³} = 4g/dm³ $$

$$ \text{Conc. of NaoH in mol/dm³} =\frac{\text{Conc. in g/dm³}}{\text{Molar mass}} $$

$$ = \frac{4}{23+16 + 1} $$ $$ = \frac{4}{40} $$ $$ = 0.1\text{ mol/dm³} $$
b (ii) Concentration of H2SO4 in mol/dm³

Since we have the concentration of the titrant in mol/dm³. Using, $$\frac{C_AV_A}{C_BV_B} = \frac{a}{b} $$ $$ \frac{C_A × 21.60}{0.1 × 25} = \frac{1}{2} $$ $$ {21.60 × C_A × 2} = {0.1 × 25} $$ $${C_A} = \frac{2.5}{43.2} $$ $$\text{Conc. of } {H_2SO_4} = {0.0579mol/dm³} $$

b(ii) Concentration of H2SO4 in g/dm³

$${\text{Conc. in g/dm³}} = {\text{Conc. in mol/dm³} × M} $$
$${\text{Molar mass, M}} = 2 + 32 + 16 × 4 $$ $${M} = 98g/mol $$ $${\text{Conc. in g/dm³}} = 0.0579 × 98 $$ $$ {\text{Conc. in g/dm³}} = 5.67g/dm³ $$

c. Precautions when carrying out titration

  1. Rinse the burette with the acid
  2. Rinse the conical flask and pipette with the base
  3. Remove the funnel from the burette before titration
  4. Expel air bubbles from the burette and pipette
  5. Avoid error due to parallax by reading the lower meniscus of the acid in the burette at eye level
  6. Place the conical flask containing the base to be titrated on a white tile or paper to detect color change
  7. Do not blow out the remaining drop of the base
  8. Use 2 to 3 drops of the indicator as indicated
  9. Ensure the burette is clamped in a straight vertical position to prevent innacuracy in burette reading
  10. Rinse all the apparatus with distilled water before use

Example 2:
A is a solution of hydrochloric acid
B is a solution containing 2.45g of anhydrous sodium trioxocarbonate(IV) in 250cm³ of solution.

  1. Put A into the burette and titrate it against 20.0cm³ or 25.0cm³ portion of B using methyl orange as indicator. Repeat the exercise to obtain consistent titres. Tabulate your burette readings and calculate the average volume of A used, the equation of the reaction involved in the titration is
$${Na_2CO_3} + {2HCl} ——> {2NaCl} + {H_2O} + {CO_2} $$
  1. From your results and the information provided. Calculate the
    1. Concentration of B in mol/dm³
    2. Concentration of A in mol/dm³
    3. Concentration of A in g/dm³
    4. Volume of the gas evolved in the reaction at S.T.P
      5. [H = 1.00, C = 12.0, Cl = 35.5, Na = 23.0, O = 16.0, Molar volume of gas at S.T.P = 22.4dm³]

Solution

At the end of the experiment the following data was obtained. All burette readings in cm³
1st titre value 2nd titre value 3rd titre value
Final burette reading 21.00 40.60 40.50
Initial burette reading 0.00 20.00 20.00
Volume of acid used 21.00 20.60 20.50
$$\text{Average titre value} = \frac{20.60 + 20.50}{2} $$ $$\text{Vol. of A used} = 20.55cm³ $$

    1. Concentration of B in mol/dm³
$$\text{If 2.45g of B } ———> {250cm³} $$ $$\text{Xg of B } ———> 1000cm³ $$ $$\text{Conc. of B in g/dm³} =\frac {1000 × 2.45}{250} $$ $$\text{Conc of B in g/dm³} = {9.8g/dm³} $$
$$\text{Conc of B in mol/dm³} = \frac{\text{Conc. of B in g/dm³}}{\text{molar mass}} $$
$$\text{Conc. of B} = \frac{9.8}{ 23 × 2 + 12 + 16 × 3}$$ $$ \text{Conc. of B}= 0.092 mol/dm³ $$
  1. Concentration of A in mol/dm³
$$\frac{C_AV_A}{C_BV_B} = \frac{a}{b} $$ $$\frac{C_A × 20.55}{0.092 × 25} = \frac{2}{1} $$ $${C_A} = \frac{4.6}{20.55} $$ $${C_A} = 0.224mol/dm³ $$
  1. Concentration of A in g/dm³
$$\text{Conc of A in mol/dm³} = \frac{\text{mass conc.}} {\text{molar mass}} $$ $$ 0.224 = \frac{\text{conc in g/dm³}}{\text{molar mass}} $$ $$\text{Molar mass of A} = 1 + 35.5 $$ $$ \text{Molar mass of A} = 36.5g/mol $$ $$\text{conc in g/dm³} = 0.224 × 36.5 $$ $$ = 8.18g/dm³ $$
  1. Volume of gas evolved at S.T.P
$$ \text{If 1 mole of }{Na_2CO_3} = \text{1 mole of }{CO_2} $$ $$ \text{1 mol} \hspace{2em} ———> 22.4dm³ $$ $$ \text{0.092 mole} \hspace{1em} ——> xdm³ $$ $$\text{Vol. of gas evolved} = 22.4 × 0.092 $$ $$\text{Vol. of gas evolved} = 2.061dm³ $$

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