■ Neutralization
► | The reaction between an acid and a base to produce a salt and water ONLY. | |||||||
► | Example
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► | H+ and OH- ions combine and mutually neutralize each other to form a neutral water molecule. | |||
► | pH value for water is 7, a neutral condition is achieved. | |||
► | In general, the neutralization process can be summarized by the following ionic equation:
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Application of neutralization in daily life
Control acidity of water
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Neutralize basic or acidic soil
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Brushing teeth
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Treat bee stings
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Prevent coagulation of latex
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Treat acidic effluent
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Acid-base titration
■ Titration
► | a quantitative analysis method | |
► | determination of the concentration of an acid or base by exactly neutralizing the acid/base with an acid or base of known concentration | |
► | with the help of a suitable indicator. |
► | neutralization occurs completely | |
► | all the OH- ions combine with all the H+ ions in the solution to form water |
► | Change colour at the end where all hydrogen ions, H+, react completely with hydroxide ion, OH-. | |||||||||||||||||||||||||
► | Types of indicators and its color transition under different acid-base conditions
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► | pH value of the solution changes sharply at the neutral point. |
► | The electrical conductivity reach its lowest point at the neutral point | |
► | All the ion dissociates from the acid and alkali have reacted completely | |
► | The water molecules formed do not dissociate into ions and thus do not conduct electricity | |
► | Example: Neutralization of sulphuric acid and barium hydroxide. |
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► | At the end point when H+ ion neutralizes OH- ion, water molecules are formed. At this point, the
electrical conductivity is zero as shown in the graph.
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► | to determine the molarity of a solution if the molarity of another solution is known | |
► | to determine the basicity of an acid | |
► | to construct a chemical equation of a neutralization reaction |
Laboratory Activity 7.4.1: Acid-base titration |
Solving problem related to neutralization reactions
■ Neutralization reaction between acidic solution A and alkaline solution B can be represented by the general equation as follows:
► | aA + bB → product | |
► | From the equation, a mol of acid A neutralize b mol of alkali. |
► | VA of MA acidic solution neutralize VB of MB alkaline solution. | |
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✍ Worked-example 7.4(a)
100cm3 of hydrochloric acid, 1 mol dm-3 react completely with 50cm3 sodium hydroxide solution, NaOH. Calculate the molarity of the sodium hydroxide solution.
Solution:
Step 1: Write a balanced chemical equation.
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
From the chemical equation, 1 mol of hydrochloric acid reacts with 1 mol of sodium hydroxide. Thus, a = 1 and b = 1
Step 2: Write the information given.
MA = 1 mol dm-3, VA = 100cm3, MB = ? mol dm-3, VB = 50cm3
Step 3: Substitute the values in the formula
MB = 2 mol dm-3
✍ Worked-example 7.4(b)
If 50cm3 of nitric acid, 0.05mol dm-3, is neutralized completely by calcium hydroxide, 0.2mol dm-3, what is the volume of the calcium hydroxide solution used?
Solution:
Step 1: Write a balanced chemical equation.
2HNO3(aq) + Ca(OH2)(aq) → Ca(NO3)2(aq) + 2H2O(l)
From the chemical equation, 2 mol of nitric acid reacts with 1 mol of calcium hydroxide. Thus, a = 2 and b = 1
Step 2: Write the information given.
MA = 0.05 mol dm-3, VA = 503, MB = 0.2 mol dm-3, VB = ? cm3
Step 3: Substitute the values in the formula
VB = 6.25cm3
Worked-example: Solving problem related to neutralization reactions
■ The following knowledge is required in the calculations involving neutralization:
► | Chemical equation or ionic equation for neutralization | |
► | Conversion of units of moles to grams or vice versa | |
► | Concept related to the concentration |
✍ Worked-example 7.4(c)
20cm3 of barium hydroxide solution completely neutralize by 10cm3 of 12.6g dm-3 nitric acid. Calculate the concentration of barium hydroxide solution in g dm-3.
[Relative atomic mass: H, 1; N, 14; O, 16; Ba, 137]
Solution:
Step 1: Write a balanced chemical equation.
2HNO3(aq) + Ba(OH2)(aq) → Ba(NO3)2(aq) + 2H2O(l)
From the chemical equation, 2 mol of nitric acid reacts with 1 mol of barium hydroxide. Thus, a = 2 and b = 1
Step 2: Write the information given.
Concentration of acids = 12.6g dm-3 MA = 0.2 mol dm-3 |
VA = 103 | MB = ? mol dm-3 | VB = 20cm3 |
MB = 0.05mol dm-3
Step 4: Unit conversion
Concentration of barium hydroxide
= 8.55g dm-3
✍ Worked-example 7.4(d)
What is the volume of 0.5mol dm-3 of sulphuric acid needed to react completely with 10cm3 sodium hydroxide solution of 40g dm-3?
Solution:
Step 1: Write a balanced chemical equation.
H2SO4(aq) + 2NaOH(aq) → Na2(SO4)(aq) + 2H2O(l)
From the chemical equation, 1 mol of sulphuric acid reacts with 2 mol of sodium hydroxide. Thus, a = 1 and b = 2
Step 2: Write the information given.
MA = 0.5 mol dm-3 | VA = ? cm3 | Relative formula mass of NaOH = 23 + 16 + 1 = 40 = 1 mol dm-3 |
VB = 10cm3 |
VB = 10cm3
⇲ For exercise(objective and subjective), download for free on Android OS. | ||
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