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SS3: CHEMISTRY - 1ST TERM

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  1. Volumetric Analysis (Titration) | Week 1
    3 Topics
    |
    1 Quiz
  2. Heat of Neutralization & Redox Titrations | Week 2
    2 Topics
    |
    1 Quiz
  3. Test for Common Gases | Week 3
    2 Topics
    |
    1 Quiz
  4. Qualitative Analysis I | Week 4
    3 Topics
    |
    1 Quiz
  5. Qualitative Analysis II | Week 5
    2 Topics
    |
    1 Quiz
  6. Qualitative Analysis III | Week 6
    3 Topics
    |
    1 Quiz
  7. Qualitative Analysis IV - Anions | Week 7
    1 Topic
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    1 Quiz
  8. Test for Fat and Oil; Protein; Starch | Week 8
    3 Topics
    |
    1 Quiz
  9. Petroleum I | Week 9
    4 Topics
    |
    1 Quiz
  10. Petroleum II | Week 10
    5 Topics
    |
    1 Quiz

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Topic Content:

  • Meaning of Water of Crystallization

What is Water of Crystallization?

Water of crystallization is an impurity in a compound. It does not take part in the chemical reaction. Hence, concentrations calculated from results during titration is that of the anhydrous substance. Water of crystallization and the percentage of water of crystallization can be calculated during titration. 

The molecules of water of crystallization contained in a hydrated salt are given by the following relations.

\( \frac{Mass\: concentration \: of \: anyhdrous\:salt}{Molar\: mass \:of \: anyhdrous\:salt} \\=\frac{Mass\: concentration \:of \: hydrated\:salt}{Molar\: mass \:of \:hydrated\:salt}\\= \frac{Mass\:concentration \:of \: water \: molecules}{Molar\: mass \:of \:water \: molecules} \)

\( \frac{Mass\:concentration\:of \: anyhdrous\:salt}{Mass\: concentration \:of \: hydrated \:salt}\\ = \frac{Molar\: mass \:of \: anyhdrous\:salt}{Molar\: mass \:of \: hydrated\:salt} \)

Mass concentration of water molecules =

Mass concentration of hydrated salt – Mass concentration of anhydrous salt

 

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Evaluation Question 1

The following table shows burette readings obtained when 25.0 cm3 of a solution of sodium trioxocarbonate(IV) containing 1.30 g in 250 cm3 of solution was titrated against a solution of hydrochloric acid of an unknown concentration:

Burette reading Rough I II
Final reading (cm3) 23.50 44.20 25.20
Initial reading (cm3) 2.00 23.50 4.30
Volume of acid used (cm3)      

 

a. Copy and complete the table.

 Burette reading Rough I II
Final reading(cm3) 23.50 44.20 25.20
Initial reading(cm3) 2.00 23.50 4.30
Volume of acid used (cm3) 21.50 20.70 20.90

 

b. Calculate the average volume of acid used.

Average volume of Acid used = \( \frac{20.70 \; + \; 20.90}{2}  \\ \scriptsize = 20.80 \: cm^3\)

c. Write a balanced equation for the reaction in the titration 

2HCl(aq) + Na2CO3(aq) → 2NaCl(aq) + H2O(l) + CO2(g)

Evaluation Question 2

In a titration experiment, 22.50 cm3 of an acid solution A containing 10.6 g of NaHSO4 per dm3 reacted with 25.0 cm3 of solution B containing X g of NaOH per dm3. The equation for the reaction is: 

NaHSO4(aq) + NaOH(aq) → Na2SO4(aq) + H2O(l) 

(a) From the information given above, calculate the:

(i) concentrate of A in mol dm– 3;

Solution

Molar conc = \( \frac{molar \; conc}{molar \; mass} \)

Molar mass of NaHSO4 = 23 + 1 + 32 + (16 × 4)

= 120 g/mol

∴ Molar conc = \(\frac{10.2}{12}\)

Answer  = 0.088 mol/dm3

(ii) concentrate of B in mol dm– 3;

where nA & nB = 1

⇒ \( \frac {C_A V_A}{C_B V_B} = \frac{n_A}{n_B} \\ \scriptsize C_B = \normalsize \frac{C_A V_A n_B}{V_B n_A} \\ \scriptsize C_B = \normalsize \frac{0.08 \; \times \; 22.5\; \times \; 1 }{25 \; \times \; 1} \\ \scriptsize = 0.0792moldm^{-3} \)

(iii) value of X;

Mass conc = molar conc × molar mass

Molar mass of NaOH = 23 + 16 + 1 = 4 0g/mol

Mass conc = 0.0792 × 40

Answer  = 3.17 g/dm3

(iv) mass of Na2SO4 formed during the reaction. 

Anhydrous salt NaHSO4.Hydrated salt = Na2SO4

⇒ \( \frac{Mass\; concentration \;of \; anyhdrous\;salt}{Mass\; concentration \;of \; hydrated \;salt} = \frac{Molar\; mass \;of \; anyhdrous\;salt}{Molar\; mass \;of \; hydrated\;salt} \)

⇒ \( \frac {120}{142} = \frac{10.6}{x} \)

x = \( \frac {142 \; \times \; 10.6}{120}  \\ \scriptsize = 12.5 g/dm^3\)

 

(b)(i) Name a suitable indicator for the titration experiment.

Answer:

Phenolphthalein


(ii) State the apparatus used to measure the volume of the solution: I. A; II B

Answer

Pipette and burette

Evaluation Question 3

Briefly explain how titration is carried out.

Answer: 25 cm3 of base is pipette into a conical flask. Then an indicator is added, the type depends on the acid/base strength. Then the acid is poured into the burette and is titrated into the base until the indicator changes colour. Then the volume of acid used is calculated.

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