7.4 Neutralization

Neutralization
■ Neutralization

	►	The reaction between an acid and a base to produce a salt and water ONLY.
	►	Example Acid + Base → Salt + water HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) 2HNO3(aq) + Mg(OH)2(aq) → Mg(NO3)2(aq) + 2 H2O(l)

■ In the neutralization process,

	►	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: ○ H+(aq) + OH- → H2O(l)

■ This video contains information on the application of neutralisation

Application of neutralization in daily life

Control acidity of water	Neutralize basic or acidic soil	Brushing teeth	Treatment of gastric with antacid.
Treat bee stings	Prevent coagulation of latex	Treat acidic effluent	Remove acidic gas

Remove acidic gas
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.

■ Neutral point[End point]

	►	neutralization occurs completely
	►	all the OH- ions combine with all the H+ ions in the solution to form water

■ Acid-base indicators

	►	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 Type of indicator Acidic solution Neutral solution Alkaline solution Universal indicator Red/orange/yellow Green Purple Litmus solution Red Purple Blue Phenolphthalein Colourless Colourless Pink Methyl orange Red Orange Yellow Methyl red Red Orange Yellow

■ pH measurement

►

pH value of the solution changes sharply at the neutral point.

■ Electrical conductivity

	►	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.
	►	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.

■ The results obtained from the titration of an acid with an alkaline solution can be used to

	►	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

■ This video contains information on the titration.

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.

■ At neutral point of neutralization reaction,

	►	VA of MA acidic solution neutralize VB of MB alkaline solution.
	►	$\text{Number of moles of acid =}\frac{M_A{V}_A}{1000}$
	►	$\text{Number of moles of alkaline =}\frac{M_B{V}_B}{1000}$
	►	$\text{Thus, the ratio of number of moles, a to number of moles of alkali b is}$ $\frac{M_A{V}_A}{M_B{V}_B}=\frac{a}{b}$

✍ 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.
M_A = 1 mol dm^-3, V_A = 100cm³, M_B = ? mol dm^-3, V_B = 50cm³
Step 3: Substitute the values in the formula
$\frac{M_A{V}_A}{M_B{V}_B}=\frac{a}{b}$
$\frac{1\mathrm{\times }100}{M_B\mathrm{\times }50}=\frac{1}{1}$
M_B = 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.
M_A = 0.05 mol dm^-3, V_A = 50³, M_B = 0.2 mol dm^-3, V_B = ? cm³
Step 3: Substitute the values in the formula
$\frac{M_A{V}_A}{M_B{V}_B}=\frac{a}{b}$
$\frac{0.05\mathrm{\times }50}{0.2\mathrm{\times }{V}_B}=\frac{2}{1}$
$V_B\text{=}\frac{0.05\mathrm{\times }50}{0.2\mathrm{\times }2}$
V_B = 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 $M_A\mathrm{=}\frac{12.6}{1+14+(\mathrm{16x3})}\mathit{mol}{\mathit{dm}}^{\mathrm{-}3}$ $M_A\mathrm{=}\frac{12.6}{63}$ MA = 0.2 mol dm-3

VA = 103

MB = ? mol dm-3

VB = 20cm3

Step 3: Substitute the values in the formula
$\frac{M_A{V}_A}{M_B{V}_B}=\frac{a}{b}$
$\frac{0.2\mathrm{\times }10}{M_B\mathrm{\times }20}=\frac{2}{1}$
$M_B\text{=}\frac{0.2\mathrm{\times }510}{2\mathrm{\times }20}$
M_B = 0.05mol dm^-3
Step 4: Unit conversion
Concentration of barium hydroxide
$\text{= 0.05}\mathrm{\times }\mathrm{[137\; +\; (1\; +\; 16)}\mathrm{\times }\mathrm{2]}$
= 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 $M_B\mathrm{=}\frac{40}{40}$ = 1 mol dm-3

VB = 10cm3

Step 3: Substitute the values in the formula
$\frac{M_A{V}_A}{M_B{V}_B}=\frac{a}{b}$
$\frac{0.5\mathrm{\times }{V}_A}{1\mathrm{\times }10}=\frac{1}{2}$
$V_B\text{=}\frac{10}{0.5\mathrm{\times }2}$
V_B = 10cm³

← 7.3 Concentration of Acid and Alkali

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