The Neutralization Reaction Between H3po4 And Naoh In An Aqueous Solution

Strength of Acid and Alkali, Reacting Masses, Volumetric AnalysisIn 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 acidA.

2B.1, 3C.3, 4D.1, 2, 4Option 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)Number of moles of H2SO4 given: 0.12 X (25.0 /1000)= 0.003molNumber of moles of NaOH given: 0.12 X (50.0/1000)= 0.006molMole ratio of H2SO4 : NaOH = 0.

003 /0.006 = 1:2Therefore, option (1) is correct.Option 2 :NaOH(aq) + NH4Cl (aq)–> NaCl(aq)+ NH3(g) + H2O(l)Mole ratio of NaOH : NH4Cl = l :1?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)Number of moles of NaOH given: 0.5X (50/1000)= 0.025molMole ratio of NaOH : NH4Cl = 0.025/ 0.025= 1:1This indicates that neither NaOH nor NH4Cl will be in excess, thus the reaction is complete.

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However, option (2) is still incorrect as the reaction between alkali and ammonium compound is not considered as neutralization. Neutralization refers to the combination of hydrogen ions, H+(aq) and hydroxide ions ,OH-(aq) ( or oxide ions,O2- ) to form water molecules, H2O(l). The product of neutralization can only be salt and water.Therefore, option (2) is incorrect.Option 3This option tests the understanding of strength of acid and alkali.Many students may have already rejected this option immediately when they have the first look on it as they have the common misconception that compared with a strong acid of the same volume and concentration, a weak acid requires a smaller amount of alkali for complete neutralization, as a result, the alkali will be in excess,thus it is not a complete neutralization. In fact, during neutralization with a strong alkali such as sodium hydroxide solution, more and more molecules of the weak acid will ionize and eventually all the molecules ionize to give H+ (aq) 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 = 1:3Using 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)= 0.002molNumber of moles of KOH given: 0.200X (30.0/1000)= 0.006molMole ratio of H3PO4: KOH = 0.002 /0.006 = 1:3Therefore, option (3) is correct.Option 4:HCl(aq) + NaOH(aq)–> NaCl(aq) +H2O(l)Using the formula,Molarity of a solution M or mol dm3= Number of moles of solute (mol) / Volume of solution (dm3)Number of moles of solute (mol)Number of moles of HCl used: 1X (50/1000)= 0.05molFrom the chemical equation,We can see that mole ratio of HCl: NaOH = 1:1Number of moles of NaOH required for complete neutralization: 1X0.05mol= 0.05molNumber of moles in 250cm3 of NaOH: Mass(g) / Molar mass(gmol-1)= 2.025/(23.5 + 16.0 + 1.0)= 0.05molNumber of moles in 25 cm3 of NaOH : 0.005/10= 0.005 molMole ratio of HCl: NaOH = 1:1Number of moles of HCl reacted with 25cm3 of NaOH: 1 X 0.005 = 0.005mol?NaoH becomes the limiting reageant (HCl is in excess)The number of moles of HCl unreacted with NaOH :0.05-0.005 = 0.045mol?The neutralization is not completeSome students may forget to divide the number of moles in 250cm3 of NaOH by 10.Consequently, they wrongly think that 0.05 mol of NaOH reacts with 0.05 mol HCl , and so the neutralization is complete. We should remember to divide the number of moles in 250cm3 of NaOH by 10 as only 25cm3 of NaOH (one-tenth of the whole NaOH solution) is used to react with HCl.Therefore, Option (4) is incorrect.Overall : Only option (1)and (3) are correct so the correct answer is B.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.