3-acid-base-titration.md

13/09/23

Aim

To prepare a Standard solution of oxalic acid and finding the molarity of the given sodium hydroxide solution.

Apparatus Required

Conical flask, burette stand, 20 ml pipette

Chemical Required

Sodium hydroxide solution, phenolphthalein indicator and oxalic acid.

Theory

The titration between oxalic acid and sodium hydroxide solution is carried out using phenolphthalein as an indicator.

Sodium hydroxide is taken in burette and oxalic acid is taken in conical flask with the help of pipette it shows pink color.

To prepare M/20 solution of Oxalic acid, the weight of oxalic acid required can be calculated by using the formula.

$$ \text{Molarity = } \frac{W_b * 1000}{M_b * \text{volume in ml} } \quad\text{or} \quad W_b = \frac{\text{Molarity} * M_b * \text{volume in ml}}{1000} $$

Where, $W_b$ = Weight of oxalic acid, $\qquad$ $M_b$ = Molecular mass of oxalic acid = 126 g/mol

Molarity = M/20

$$ W_b = \frac{126 * 250}{20*1000} $$

$$ \implies \text{1.57 g} $$

Weight of Oxalic acid required = 1.57 g

Reaction involved

Molecular equation = $\underset{|\underset{COOH}{}}{\text{COOH}}$ $\cdot$ $\text{NaOH}$ $\rightarrow$ $(\text{COONa})$ + $2\text{H}_2\text{O}$

Don't write this and the next sentence. Refer to page 40 of Chemistry Lab manual for _correctly rendered _equations.__

Observation

Volume of pipette = 10 ml Molarity of the Oxalic acid solution = 1/20 mol/liter
Weight of Oxalic acid dissolved in 250 ml of distilled water = 1.57 g

S.No	Volume of Oxalic Acid solution	Initial Reading of burette	Final reading of burette	Difference
1.	10 ml	0	39.6	39.6
2.	10 ml	0	39.5	39.5
3.	10 ml	0	39.6	39.6
4.	10 ml	0	39.7	39.7
5.	10 ml	0	39.6	39.6

Calculations

$$ \text{Oxalic Acid vs. NaOH} $$

$$ \underset{(\text{Oxalic acid})}{n_1M_1V_1}\quad=\quad\underset{\text{(NaOH)}}{n_2M_2V_2} $$

As per ionic equation numbers of electrons lost or gained $n_1 = 2$ and $n_2 = 1$

$$ 2 * M_1V_1 = 1 * M_2V_2 $$

$$ M_2 = \frac{2}{1} \cdot \frac{M_1V_1}{V_2} $$

Molecular mass of NaOH = 40
Molarity of NaOH solution = $M_2$
Strength of NaOH solution = $M_2$*40 g = $\underline{xyz}$ g/liter

Precautions

1. Always add NaOH solution drop and take the reading of NaOH solution in the burette from the lover meniscus.
2. Always use distilled water for preparing the solution.
3. Last drop of the solution should not be blown out of the pipette.
4. End point should be determined properly.

Results

Molarity of the given NaOH solution = $\underline{xyz}$ M
Strength of given NaOH solution = $\underline{xyz}$ g/l