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The table for burette readings is empty. To complete the calculations, I will assume an average titre value for the acid.

Chemistry

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The table for burette readings is empty. To complete the calculations, I will assume an average titre value for the acid.

Assumption:
Let the average titre value (volume of acid used) be 22.50 cm3\text{22.50 cm}^3.

Given Information:

  • Solution B: Contains 0.02 mol of sodium hydroxide (NaOH).
  • Volume of solution B (stock) = 250 cm³.
  • Solution A: Tetraoxosulphate (IV) acid (H2SO4H_2SO_4).
  • Volume of solution B used in titration (VbV_b) = 25 cm³.
  • Average titre value (volume of acid A used, VaV_a) = 22.50 cm³ (assumed).

Step 1: Calculate the concentration of the base (Solution B).
Moles of NaOH = 0.02 mol
Volume of stock solution B = 250 cm³ = 0.250 dm³

Mb=Moles of NaOHVolume of solution B (dm3)=0.02 mol0.250 dm3=0.08 mol/dm3M_b = \frac{\text{Moles of NaOH}}{\text{Volume of solution B (dm}^3)} = \frac{0.02 \text{ mol}}{0.250 \text{ dm}^3} = 0.08 \text{ mol/dm}^3

So, the molar concentration of the base (MbM_b) is 0.08 mol/dm30.08 \text{ mol/dm}^3.

Step 2: Write the balanced chemical equation for the reaction.
The reaction is between sulfuric acid (H2SO4H_2SO_4) and sodium hydroxide (NaOH).

H2SO4(aq)+2NaOH(aq)→Na2SO4(aq)+2H2O(l)\text{H}_2\text{SO}_4 (aq) + 2\text{NaOH} (aq) \rightarrow \text{Na}_2\text{SO}_4 (aq) + 2\text{H}_2\text{O} (l)

From the balanced equation:

  • Moles of acid (nan_a) = 1
  • Moles of base (nbn_b) = 2

Part i· 2 steps

**The molar concentration of the acid.

  1. Convert volumes to dm³.

    Va=22.50 cm3=0.02250 dm3V_a = 22.50 \text{ cm}^3 = 0.02250 \text{ dm}^3
    Vb=25 cm3=0.025 dm3V_b = 25 \text{ cm}^3 = 0.025 \text{ dm}^3

  2. Use the titration formula.

    MaVana=MbVbnb\frac{M_aV_a}{n_a} = \frac{M_bV_b}{n_b}

    Rearrange to solve for MaM_a:

    Ma=MbVbnaVanbM_a = \frac{M_bV_b n_a}{V_a n_b}

    Substitute the known values:

    Ma=(0.08 mol/dm3)×(0.025 dm3)×1(0.02250 dm3)×2M_a = \frac{(0.08 \text{ mol/dm}^3) \times (0.025 \text{ dm}^3) \times 1}{(0.02250 \text{ dm}^3) \times 2} Ma=0.0020.045M_a = \frac{0.002}{0.045} Ma=0.04444... mol/dm3M_a = 0.04444... \text{ mol/dm}^3

    Rounding to three significant figures:

    Ma=0.0444 mol/dm3M_a = \boxed{\text{0.0444 mol/dm}^3}

Part ii· 2 steps

**The concentration mass of the acid.

  1. Calculate the molar mass of H2SO4H_2SO_4.

    Atomic masses: H = 1.01 g/mol, S = 32.07 g/mol, O = 16.00 g/mol

    MMH2SO4=(2×1.01)+32.07+(4×16.00)MM_{H_2SO_4} = (2 \times 1.01) + 32.07 + (4 \times 16.00) MMH2SO4=2.02+32.07+64.00=98.09 g/molMM_{H_2SO_4} = 2.02 + 32.07 + 64.00 = 98.09 \text{ g/mol}

  2. Calculate the mass concentration.

    Mass Concentration=Ma×MMH2SO4\text{Mass Concentration} = M_a \times MM_{H_2SO_4} Mass Concentration=0.04444 mol/dm3×98.09 g/mol\text{Mass Concentration} = 0.04444 \text{ mol/dm}^3 \times 98.09 \text{ g/mol} Mass Concentration=4.359 g/dm3\text{Mass Concentration} = 4.359 \text{ g/dm}^3

    Rounding to three significant figures:

    Mass Concentration=4.36 g/dm3\text{Mass Concentration} = \boxed{\text{4.36 g/dm}^3}

Part iii· 2 steps

**From a balanced chemical equation, determine the mole ratio of the base.

  1. Write the balanced chemical equation.

    H2SO4(aq)+2NaOH(aq)→Na2SO4(aq)+2H2O(l)\text{H}_2\text{SO}_4 (aq) + 2\text{NaOH} (aq) \rightarrow \text{Na}_2\text{SO}_4 (aq) + 2\text{H}_2\text{O} (l)

  2. Determine the mole ratio.

    The stoichiometric coefficient for the base (NaOH) is 2.
    The mole ratio of acid to base (H2SO4:NaOHH_2SO_4 : NaOH) is 1:21:2.
    The mole ratio of the base (NaOH) in the reaction is 2\boxed{2}.

Part iv

**Sketch the titration curve of the acid-base titration against the pH.

This is a titration of a strong base (NaOH) with a strong acid (H2SO4H_2SO_4). The base is initially in the conical flask, and the acid is added from the burette.

Characteristics of the curve:

  • Initial pH: High (basic), as the solution starts with NaOH. Initial pH of 0.08 M NaOH: pOH=−log⁡(0.08)=1.10pOH = -\log(0.08) = 1.10, so pH=14−1.10=12.90pH = 14 - 1.10 = 12.90.
  • Equivalence point: At pH 7.0, because it's a strong acid-strong base titration.
  • Shape: The pH decreases gradually, then sharply drops around the equivalence point (pH 7), and then continues to decrease gradually as excess acid is added.
  • Final pH: Low (acidic), as excess H2SO4H_2SO_4 is added.

Here is a sketch of the titration curve:

\begin{tikzpicture}[scale=0.8] \draw[->] (0,0) -- (10,0) node[below] {Volume of Acid (cm$^3$)}; \draw[->] (0,0) -- (0,10) node
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