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A titration experiment involves solutions F (H₂X) and G (NaOH). Pipette volume is 25 cm³. Use the teacher's hint that average burette reading for F is 24.50 cm³. Calculate the concentration of G (NaOH

Chemistry

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3 parts
b(i)

0.100 mol dm−3\text{0.100 mol dm}^{-3}

b(ii)

0.0510 mol dm−3\text{0.0510 mol dm}^{-3}

b(iii)

95.1 g/mol\text{95.1 g/mol}

Step-by-step solution

5 steps

Here are the calculations based on the provided information:

a) The question asks to record the pipette volume, tabulate burette readings, and calculate the average volume of F used. Since no experimental data is provided, we will use the information given in the problem and the teacher's hint.

The volume of the pipette used for solution G (NaOH) is assumed to be 25 cm325 \text{ cm}^3.
The average volume of F (H₂X) used from the burette, as suggested by the teacher's hint, is 24.50 cm324.50 \text{ cm}^3.

Part b(i)· 3 steps

Concentration of G in mol dm−3\text{mol dm}^{-3}.

Solution G contains 1.00 g1.00 \text{ g} of sodium hydroxide (NaOH) in 250 cm3250 \text{ cm}^3 of solution.
The atomic masses are H = 1, O = 16, Na = 23.

  1. Calculate the molar mass of NaOH.

    Molar mass of NaOH=Na+O+H=23+16+1=40 g/mol\text{Molar mass of NaOH} = \text{Na} + \text{O} + \text{H} = 23 + 16 + 1 = 40 \text{ g/mol}

  2. Calculate the number of moles of NaOH.

    Moles of NaOH=Mass of NaOHMolar mass of NaOH=1.00 g40 g/mol=0.025 mol\text{Moles of NaOH} = \frac{\text{Mass of NaOH}}{\text{Molar mass of NaOH}} = \frac{1.00 \text{ g}}{40 \text{ g/mol}} = 0.025 \text{ mol}

  3. Calculate the concentration of NaOH in mol dm−3\text{mol dm}^{-3}.

    The volume of the solution is 250 cm3250 \text{ cm}^3. Convert this to dm3\text{dm}^3:

    250 cm3=2501000 dm3=0.250 dm3250 \text{ cm}^3 = \frac{250}{1000} \text{ dm}^3 = 0.250 \text{ dm}^3 Concentration of G (NaOH)=Moles of NaOHVolume of solution in dm3=0.025 mol0.250 dm3=0.100 mol dm−3\text{Concentration of G (NaOH)} = \frac{\text{Moles of NaOH}}{\text{Volume of solution in dm}^3} = \frac{0.025 \text{ mol}}{0.250 \text{ dm}^3} = 0.100 \text{ mol dm}^{-3}

    The concentration of G is 0.100 mol dm−3\boxed{\text{0.100 mol dm}^{-3}}.

Part b(ii)

0.0510 mol dm−3\text{0.0510 mol dm}^{-3}

Concentration of F in mol dm−3\text{mol dm}^{-3}.

The balanced chemical equation is:

H2X(aq)+2NaOH(aq)→Na2X(aq)+2H2O(l)\text{H}_2\text{X(aq)} + 2\text{NaOH(aq)} \rightarrow \text{Na}_2\text{X(aq)} + 2\text{H}_2\text{O(l)}

From the equation, the mole ratio of H₂X to NaOH is 1:21:2.

Given values:
Concentration of base, Cb=Concentration of G=0.100 mol dm−3C_b = \text{Concentration of G} = 0.100 \text{ mol dm}^{-3}
Volume of base, Vb=Pipette volume of G=25 cm3=0.025 dm3V_b = \text{Pipette volume of G} = 25 \text{ cm}^3 = 0.025 \text{ dm}^3
Volume of acid, Va=Average volume of F=24.50 cm3=0.0245 dm3V_a = \text{Average volume of F} = 24.50 \text{ cm}^3 = 0.0245 \text{ dm}^3
Moles of acid from equation, na=1n_a = 1
Moles of base from equation, nb=2n_b = 2

Using the titration formula:

CaVaCbVb=nanb\frac{C_a V_a}{C_b V_b} = \frac{n_a}{n_b}

Rearrange to solve for CaC_a:

Ca=nanb×CbVbVaC_a = \frac{n_a}{n_b} \times \frac{C_b V_b}{V_a}

Substitute the values:

Ca=12×(0.100 mol dm−3)×(0.025 dm3)0.0245 dm3C_a = \frac{1}{2} \times \frac{(0.100 \text{ mol dm}^{-3}) \times (0.025 \text{ dm}^3)}{0.0245 \text{ dm}^3} Ca=12×0.00250.0245C_a = \frac{1}{2} \times \frac{0.0025}{0.0245} Ca=12×0.1020408...C_a = \frac{1}{2} \times 0.1020408... Ca=0.0510204... mol dm−3C_a = 0.0510204... \text{ mol dm}^{-3}

Rounding to three significant figures:

Ca=0.0510 mol dm−3C_a = 0.0510 \text{ mol dm}^{-3}

The concentration of F is 0.0510 mol dm−3\boxed{\text{0.0510 mol dm}^{-3}}.

Part b(iii)

95.1 g/mol\text{95.1 g/mol}

Molar mass of the acid H₂X.

We are given that 100 cm3100 \text{ cm}^3 of solution F contained 0.485 g0.485 \text{ g} of acid.
From part (b)(ii), the concentration of F is 0.0510 mol dm−30.0510 \text{ mol dm}^{-3}.

  1. Calculate the number of moles of H₂X in 100 cm3100 \text{ cm}^3 of solution F.

    Convert 100 cm3100 \text{ cm}^3 to dm3\text{dm}^3:

    100 cm3=1001000 dm3=0.100 dm3100 \text{ cm}^3 = \frac{100}{1000} \text{ dm}^3 = 0.100 \text{ dm}^3 Moles of H2X=Concentration×Volume in dm3\text{Moles of H}_2\text{X} = \text{Concentration} \times \text{Volume in dm}^3 Moles of H2X=0.0510204 mol dm−3×0.100 dm3=0.00510204 mol\text{Moles of H}_2\text{X} = 0.0510204 \text{ mol dm}^{-3} \times 0.100 \text{ dm}^3 = 0.00510204 \text{ mol}

  2. Calculate the molar mass of H₂X.

    Molar mass of H2X=Mass of H2XMoles of H2X\text{Molar mass of H}_2\text{X} = \frac{\text{Mass of H}_2\text{X}}{\text{Moles of H}_2\text{X}} Molar mass of H2X=0.485 g0.00510204 mol=95.059 g/mol\text{Molar mass of H}_2\text{X} = \frac{0.485 \text{ g}}{0.00510204 \text{ mol}} = 95.059 \text{ g/mol}

    Rounding to three significant figures:

    Molar mass of H2X=95.1 g/mol\text{Molar mass of H}_2\text{X} = 95.1 \text{ g/mol}

    The molar mass of the acid H₂X is 95.1 g/mol\boxed{\text{95.1 g/mol}}.

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