Question

250 mL of a solution containing 0.6625 g of sodium carbonate and 2.8000 g of potassium hydroxide is titrated against N/10HCl. Find out the value of x-y, where x and y are titre value (in mL) with methyl orange and phenolphthalein, respectively, taking 20 mL of above solution each time. (Given: Atomic mass in amu of Na = 23, C = 12, O = 16, K = 39, H = 1, Cl = 35.5)

Answer 1

5

Answer 2

The solution contains sodium carbonate ($\text{Na}_2\text{CO}_3$) and potassium hydroxide ($\text{KOH}$). This is a mixture of a weak base and a strong base. We need to determine the titre values with phenolphthalein and methyl orange indicators.

1. Calculate Moles of Components in the 20 mL Aliquot:

• Molar mass of $\text{Na}_2\text{CO}_3$: $2(23) + 12 + 3(16) = 46 + 12 + 48 = 106 \text{ g/mol}$

• Molar mass of $\text{KOH}$: $39 + 16 + 1 = 56 \text{ g/mol}$

• Moles of $\text{Na}_2\text{CO}_3$ in 250 mL solution: $\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{0.6625 \text{ g}}{106 \text{ g/mol}} = 0.00625 \text{ mol}$

• Moles of $\text{KOH}$ in 250 mL solution: $\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{2.8000 \text{ g}}{56 \text{ g/mol}} = 0.05 \text{ mol}$

• Moles of $\text{Na}_2\text{CO}_3$ in 20 mL aliquot: $\text{Moles}_{20 \text{ mL}} = 0.00625 \text{ mol} \times \frac{20 \text{ mL}}{250 \text{ mL}} = 0.00625 \times 0.08 = 0.0005 \text{ mol}$

• Moles of $\text{KOH}$ in 20 mL aliquot: $\text{Moles}_{20 \text{ mL}} = 0.05 \text{ mol} \times \frac{20 \text{ mL}}{250 \text{ mL}} = 0.05 \times 0.08 = 0.004 \text{ mol}$

The titrant is $\text{N/10 HCl}$, which means its normality is $0.1 \text{ N}$.

2. Titration with Phenolphthalein Indicator (Titre value = y mL):

Phenolphthalein changes color when the solution pH is around 8.2-10. This indicates the completion of the following reactions:

• Strong base ($\text{KOH}$) reacts completely: $\text{KOH} + \text{HCl} \rightarrow \text{KCl} + \text{H}_2\text{O}$ (1 mole $\text{KOH}$ consumes 1 mole $\text{HCl}$, n-factor for $\text{KOH}$ is 1)
• Weak base ($\text{Na}_2\text{CO}_3$) reacts partially to form bicarbonate: $\text{Na}_2\text{CO}_3 + \text{HCl} \rightarrow \text{NaHCO}_3 + \text{NaCl}$ (1 mole $\text{Na}_2\text{CO}_3$ consumes 1 mole $\text{HCl}$, n-factor for $\text{Na}_2\text{CO}_3$ in this step is 1)

Total equivalents of $\text{HCl}$ required for phenolphthalein endpoint: $\text{Equivalents}_{\text{phenolphthalein}} = (\text{Moles of KOH} \times 1) + (\text{Moles of Na}_2\text{CO}_3 \times 1)$ $\text{Equivalents}_{\text{phenolphthalein}} = (0.004 \text{ mol} \times 1) + (0.0005 \text{ mol} \times 1)$ $\text{Equivalents}_{\text{phenolphthalein}} = 0.004 + 0.0005 = 0.0045 \text{ eq}$

Volume of $\text{HCl}$ (y) = $\frac{\text{Equivalents}}{\text{Normality of HCl}} = \frac{0.0045 \text{ eq}}{0.1 \text{ N}} = 0.045 \text{ L} = 45 \text{ mL}$ So, $\text{y} = 45 \text{ mL}$.

3. Titration with Methyl Orange Indicator (Titre value = x mL):

Methyl orange changes color when the solution pH is around 3.1-4.4. This indicates the completion of the following reactions:

• Strong base ($\text{KOH}$) reacts completely: $\text{KOH} + \text{HCl} \rightarrow \text{KCl} + \text{H}_2\text{O}$ (1 mole $\text{KOH}$ consumes 1 mole $\text{HCl}$, n-factor for $\text{KOH}$ is 1)
• Weak base ($\text{Na}_2\text{CO}_3$) reacts completely to form carbonic acid: $\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow \text{H}_2\text{CO}_3 + 2\text{NaCl}$ (1 mole $\text{Na}_2\text{CO}_3$ consumes 2 moles $\text{HCl}$, n-factor for $\text{Na}_2\text{CO}_3$ in this step is 2)

Total equivalents of $\text{HCl}$ required for methyl orange endpoint: $\text{Equivalents}_{\text{methyl orange}} = (\text{Moles of KOH} \times 1) + (\text{Moles of Na}_2\text{CO}_3 \times 2)$ $\text{Equivalents}_{\text{methyl orange}} = (0.004 \text{ mol} \times 1) + (0.0005 \text{ mol} \times 2)$ $\text{Equivalents}_{\text{methyl orange}} = 0.004 + 0.001 = 0.0050 \text{ eq}$

Volume of $\text{HCl}$ (x) = $\frac{\text{Equivalents}}{\text{Normality of HCl}} = \frac{0.0050 \text{ eq}}{0.1 \text{ N}} = 0.050 \text{ L} = 50 \text{ mL}$ So, $\text{x} = 50 \text{ mL}$.

4. Calculate x - y: $\text{x - y} = 50 \text{ mL} - 45 \text{ mL} = 5 \text{ mL}$

The value of x - y is 5 mL.