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Stoichiometry - Concentration and titrations

Grade 11IGCSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Concentration measures the amount of solute dissolved in a specific volume of solution, typically expressed in mol/dm3mol/dm^3 (molarity) or g/dm3g/dm^3.

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To convert volume from cm3cm^3 to dm3dm^3, divide by 10001000 (e.g., 1000cm3=1dm31000 cm^3 = 1 dm^3).

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A standard solution is a solution whose concentration is accurately known.

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Titration is a laboratory technique used to determine the concentration of an unknown solution by reacting it with a standard solution.

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The end point of a titration is indicated by a color change in a chemical indicator, such as phenolphthalein (pink in alkali, colorless in acid) or methyl orange (yellow in alkali, red in acid).

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Stoichiometry in titrations requires a balanced chemical equation to determine the mole ratio between the reactants.

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The equivalence point is the theoretical point where the moles of acid and base are in the exact ratio defined by the balanced equation.

📐Formulae

n=C×Vn = C \times V

C(mol/dm3)=n(mol)V(dm3)C (mol/dm^3) = \frac{n (mol)}{V (dm^3)}

C(g/dm3)=C(mol/dm3)×MrC (g/dm^3) = C (mol/dm^3) \times M_r

V(dm3)=V(cm3)1000V(dm^3) = \frac{V(cm^3)}{1000}

C1V1n1=C2V2n2\frac{C_1 V_1}{n_1} = \frac{C_2 V_2}{n_2}

💡Examples

Problem 1:

Calculate the concentration in mol/dm3mol/dm^3 of a solution containing 4.0g4.0 g of NaOHNaOH dissolved in 250cm3250 cm^3 of water. (MrM_r of NaOH=40NaOH = 40)

Solution:

n=massMr=4.040=0.1moln = \frac{mass}{M_r} = \frac{4.0}{40} = 0.1 mol V=2501000=0.25dm3V = \frac{250}{1000} = 0.25 dm^3 C=nV=0.10.25=0.4mol/dm3C = \frac{n}{V} = \frac{0.1}{0.25} = 0.4 mol/dm^3

Explanation:

First, convert the mass of NaOHNaOH to moles using the MrM_r. Then, convert the volume from cm3cm^3 to dm3dm^3. Finally, divide the moles by the volume to find the molarity.

Problem 2:

In a titration, 25.0cm325.0 cm^3 of 0.10mol/dm30.10 mol/dm^3 HClHCl reacts with 20.0cm320.0 cm^3 of NaOHNaOH solution. Calculate the concentration of the NaOHNaOH. The equation is: HCl(aq)+NaOH(aq)→NaCl(aq)+H2O(l)HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H_2O(l).

Solution:

n(HCl)=C×V=0.10×25.01000=0.0025moln(HCl) = C \times V = 0.10 \times \frac{25.0}{1000} = 0.0025 mol Ratio HCl:NaOHHCl:NaOH is 1:11:1, so n(NaOH)=0.0025moln(NaOH) = 0.0025 mol. C(NaOH)=nV=0.00250.020=0.125mol/dm3C(NaOH) = \frac{n}{V} = \frac{0.0025}{0.020} = 0.125 mol/dm^3

Explanation:

Calculate the moles of the known substance (HClHCl). Use the stoichiometric ratio from the balanced equation (1:11:1) to find the moles of NaOHNaOH. Divide these moles by the volume of NaOHNaOH (converted to dm3dm^3) to find its concentration.

Problem 3:

A student titrates 25.0cm325.0 cm^3 of 0.050mol/dm30.050 mol/dm^3 H2SO4H_2SO_4 against KOHKOH. If 30.0cm330.0 cm^3 of KOHKOH is required, find the concentration of KOHKOH. Equation: H2SO4+2KOH→K2SO4+2H2OH_2SO_4 + 2KOH \rightarrow K_2SO_4 + 2H_2O.

Solution:

n(H2SO4)=0.050×0.025=0.00125moln(H_2SO_4) = 0.050 \times 0.025 = 0.00125 mol Mole ratio H2SO4:KOH=1:2H_2SO_4:KOH = 1:2. n(KOH)=0.00125×2=0.0025moln(KOH) = 0.00125 \times 2 = 0.0025 mol C(KOH)=0.00250.030=0.0833mol/dm3C(KOH) = \frac{0.0025}{0.030} = 0.0833 mol/dm^3

Explanation:

Identify the mole ratio from the balanced equation. Because 11 mole of H2SO4H_2SO_4 reacts with 22 moles of KOHKOH, we multiply the moles of acid by 22 before calculating the concentration of the base.

Concentration and titrations - Revision Notes & Key Formulas | IGCSE Grade 11 Chemistry