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Chem MC analysis. In which of the following cases may it obtain a complete neutralization? (1)25.0 cm3 of 0.120 M sulphuric acid and 50cm3of 0.120M sodium hydroxide solution (2)50.cm3 of 0.5 M Sodium hydroxide and 0.025 moles of aqueous ammonium chlorid

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Introduction

Chemistry 4A Ivan Liu Chun Pok (12) Non-practical task Mc analysis Topics: Neutralization, Reaction between alkali and ammonium compound, Strength of Acid and Alkali, Reacting Masses, Volumetric Analysis In which of the following cases may it obtain a complete neutralization? (1)25.0 cm3 of 0.120 M sulphuric acid and 50cm3of 0.120M sodium hydroxide solution (2)50.cm3 of 0.5 M Sodium hydroxide and 0.025 moles of aqueous ammonium chloride (3)20.0cm3 of 0.100M phosphoric acid and 30.0cm3 of 0.200 M potassium hydroxide solution (4)Dissolve 0.2025g of solid sodium hydroxide in water and make up to 250cm3 of solution, then 25.0cm3 of this solution is added to 50.0cm3 of 1M hydrochloric acid A.2 B.1, 3 C.3, 4 D.1, 2, 4 Option 1: H2SO4(aq) + 2NaOH(aq)--> Na2SO4(aq) +H2O(l) Mole ratio of H2SO4 : NaOH = 1:2 ?Using the formula, Molarity of a solution M or mol dm-3 = Number of moles of solute (mol) / Volume of solution (dm3) ?Number of moles of solute (mol) = Molarity of a solution M or mol dm-3X Volume of solution (dm3) ...read more.

Middle

ions. Also, they may over-focus on the concept that weak acid can react with strong alkali to produce acidic salt, instead of normal salt. e.g : H3PO4(aq) + NaOH(aq)--> NaH2PO4(aq) +H2O(l) H3PO4(aq) + 2NaOH(aq)--> Na2HPO4(aq) +2H2O(l) These above reactions are absolutely correct, but we have to answer carefully for what the question asks.It said 'In which of the following cases may it obtain a complete neutralization?' It is not necessary for the salts produced is acidic, we cannot eliminate the case that normal salt is formed. Whether the neutralization is complete or not can only be determined by mole calculation. H3PO4(aq) + 3KOH(aq)--> K3PO4(aq) +3H2O(l) Mole ratio of H3PO4: KOH = 1:3 Using the formula, ? Molarity of a solution M or mol dm-3 = Number of moles of solute (mol) / Volume of solution (dm3) ?Number of moles of solute (mol) = Molarity of a solution M or mol dm-3 X Volume of solution (dm3) Number of moles of H3PO4 given: 0.100 X (20.0 /1000) ...read more.

Conclusion

If a student chooses A, it means he has wrong concepts on mole calculation (reacting masses and volumetric analysis), neutralization and strength of acid and alkali. If a student chooses C, it means he has wrong concepts on mole calculation (reacting masses and volumetric analysis). If a student chooses D, it means he has wrong concepts on strength of acid and alkali. To conclude, The steps to tackle this question are as follows: 1) See if there are reactions that are not considered as neutralization first. Option (2) can be eliminated instantly, thus the correct answer can only either be B or C. 2) In both answers B and C, we can see that option (3) is included so it implies that options (3) must be correct. It is unnecessary to determine whether it is correct or not by mole calculation. 3) We only need to determinate whether options (1) and (4) is correct or not based on mole calculation. If the calculation is correct, we can deduce that option (1) is correctwhereas option (4) is incorrect. 4) Finally we can conclude that only options (1) and (3) are correct, which lead to the fact that the correct answer should be B. ...read more.

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Clear, step by step calculations and analyses of the four reactions. The author demonstrates a detailed understanding of the chemistry involved and possible pitfalls.

Marked by teacher Adam Roberts 14/10/2013

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