Calculate the original concentration of NaOH solution A given titration data with oxalic acid and a monobasic acid.

Here's a breakdown of the calculations for your titration experiment.

Given Information:

• Solution A: Aqueous NaOH (original concentration unknown)
• Solution B: 0.05 mol/dm$^3$ oxalic acid (H$_2$C$_2$O$_4$)
• Solution C: 0.1 mol/dm$^3$ of a monobasic acid
• Dilution: 25.0 cm$^3$ of original Solution A diluted to 250 cm$^3$ (diluted Solution A).

Titration Data:

• Table 1 (Solution B vs. diluted Solution A):
  • Volume of Solution B (oxalic acid) used = 25.0 cm$^3$
  • Average titre (t$_1$) of diluted Solution A (NaOH) = 25.4 cm$^3$
• Table 2 (Solution C vs. diluted Solution A):
  • Volume of Solution C (monobasic acid) used = 25.0 cm$^3$
  • Average titre (t$_2$) of diluted Solution A (NaOH) = 13.0 cm$^3$

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Calculations:

a) The concentration of diluted solution A using the titre, t$_1$.

Step 1: Write the balanced chemical equation for the reaction between oxalic acid (H$_2$C$_2$O$_4$) and sodium hydroxide (NaOH).
Oxalic acid is a dibasic acid, meaning it donates two protons.

$$\text{H}_2\text{C}_2\text{O}_4\text{(aq)} + 2\text{NaOH(aq)} \longrightarrow \text{Na}_2\text{C}_2\text{O}_4\text{(aq)} + 2\text{H}_2\text{O(l)}$$

The mole ratio of H$_2$C$_2$O$_4$ : NaOH is $1:2$.

Step 2: Calculate the moles of oxalic acid (Solution B) used.
Volume of Solution B = $25.0 \text{ cm}^3 = 0.025 \text{ dm}^3$
Concentration of Solution B = $0.05 \text{ mol/dm}^3$

$$\text{Moles of H}_2\text{C}_2\text{O}_4 = \text{Concentration} \times \text{Volume}$$ $$\text{Moles of H}_2\text{C}_2\text{O}_4 = 0.05 \text{ mol/dm}^3 \times 0.025 \text{ dm}^3 = 0.00125 \text{ mol}$$

Step 3: Calculate the moles of NaOH (diluted solution A) that reacted.
Using the mole ratio from Step 1 (1 mole H$_2$C$_2$O$_4$ reacts with 2 moles NaOH):

$$\text{Moles of NaOH} = 2 \times \text{Moles of H}_2\text{C}_2\text{O}_4$$ $$\text{Moles of NaOH} = 2 \times 0.00125 \text{ mol} = 0.0025 \text{ mol}$$

Step 4: Calculate the concentration of diluted solution A (NaOH).
Volume of diluted Solution A used (t$_1$) = $25.4 \text{ cm}^3 = 0.0254 \text{ dm}^3$

$$\text{Concentration of diluted Solution A} = \frac{\text{Moles of NaOH}}{\text{Volume of diluted Solution A}}$$ $$$$