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Chemistry: Environmental Systems - Water Purity, Impurities, and Chemical Testing

Grade 8IB

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

🔑Concepts

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Water Purity: Pure water is a colorless, odorless liquid with a fixed melting point of 0∘C0^\circ C and a boiling point of 100∘C100^\circ C at 11 atmosphere of pressure. Impurities like dissolved salts increase the boiling point and decrease the melting point.

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Chemical Tests for Water: To identify the presence of water, we use anhydrous copper(II) sulfate (CuSO4CuSO_4), which changes from white to blue, or anhydrous cobalt(II) chloride (CoCl2CoCl_2), which changes from blue to pink.

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Soluble and Insoluble Impurities: Water can contain insoluble impurities like sand and clay (removed by filtration) and soluble impurities like minerals (Na+Na^+, Ca2+Ca^{2+}, Mg2+Mg^{2+}) and bacteria (treated with chlorine Cl2Cl_2).

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Water Treatment Process: The standard treatment involve several steps: Sedimentation (settling of large particles), Filtration (passing through sand/gravel), and Chlorination (adding Cl2Cl_2 to kill microbes).

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Hard Water vs. Soft Water: Hard water contains high concentrations of dissolved Ca2+Ca^{2+} and Mg2+Mg^{2+} ions. It reacts with soap to form 'scum' rather than 'lather'.

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Distillation: A process used to obtain pure water from a solution by evaporating the water and then condensing the steam back into a liquid, leaving non-volatile impurities behind.

📐Formulae

CuSO4(s)+5H2O(l)→CuSO4⋅5H2O(s)CuSO_4(s) + 5H_2O(l) \rightarrow CuSO_4 \cdot 5H_2O(s) (White to Blue)

CoCl2(s)+6H2O(l)→CoCl2⋅6H2O(s)CoCl_2(s) + 6H_2O(l) \rightarrow CoCl_2 \cdot 6H_2O(s) (Blue to Pink)

Ca(HCO3)2(aq)→heatCaCO3(s)+H2O(l)+CO2(g)Ca(HCO_3)_2(aq) \xrightarrow{\text{heat}} CaCO_3(s) + H_2O(l) + CO_2(g) (Removing temporary hardness)

Purity (%)=Mass of pure substanceTotal mass of sample×100\text{Purity (\%)} = \frac{\text{Mass of pure substance}}{\text{Total mass of sample}} \times 100

💡Examples

Problem 1:

A sample of water is heated and found to boil at 101.5∘C101.5^\circ C. Determine if the water is pure and explain why.

Solution:

The water is not pure.

Explanation:

Pure water has a specific boiling point of exactly 100∘C100^\circ C at 1 atm1\text{ atm}. The presence of dissolved impurities, such as salts, causes 'boiling point elevation', meaning the liquid boils at a temperature higher than 100∘C100^\circ C.

Problem 2:

Calculate the mass of water required to fully hydrate 159.6 g159.6\text{ g} of anhydrous copper(II) sulfate (CuSO4CuSO_4) to form blue crystals (CuSO4⋅5H2OCuSO_4 \cdot 5H_2O). Use Molar Mass: CuSO4=159.6 g/molCuSO_4 = 159.6\text{ g/mol} and H2O=18.0 g/molH_2O = 18.0\text{ g/mol}.

Solution:

159.6 g CuSO4+90.0 g H2O→249.6 g CuSO4⋅5H2O159.6\text{ g } CuSO_4 + 90.0\text{ g } H_2O \rightarrow 249.6\text{ g } CuSO_4 \cdot 5H_2O

Explanation:

According to the formula CuSO4⋅5H2OCuSO_4 \cdot 5H_2O, 1 mole1\text{ mole} of CuSO4CuSO_4 reacts with 5 moles5\text{ moles} of H2OH_2O. Since 1 mole1\text{ mole} of CuSO4CuSO_4 is 159.6 g159.6\text{ g}, we need 5×18.0=90.0 g5 \times 18.0 = 90.0\text{ g} of water.

Problem 3:

A student measures the mass of a water sample as 50 g50\text{ g}. After evaporating the water completely, a solid residue of 0.5 g0.5\text{ g} remains. What is the percentage of impurities in the sample?

Solution:

0.5÷50=0.010.01×100=1%\begin{array}{r} 0.5 \div 50 = 0.01 \\ 0.01 \times 100 = 1\% \end{array}

Explanation:

The mass of impurities is 0.5 g0.5\text{ g} out of a total mass of 50 g50\text{ g}. Using the formula mass of impuritytotal mass×100\frac{\text{mass of impurity}}{\text{total mass}} \times 100, we get 1%1\% impurity concentration.