Strong Bases

Strong Bases Definition

According to Arrhenius, bases are compounds that releases hydroxide ions OH in aqueous solution. Strong bases are those bases which can dissociates completely to give hydroxide ions in aqueous solution. Such as,

KOH(aq) → K+(aq) + OH(aq)

That means, one mole of strong base dissociates in aqueous solution to give one mole of hydroxide ion and one mole of positive ion. Alkali and alkaline earth metals or group I and group II elements are usually form strong bases.

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Such as:

  • Lithium hydroxide- LiOH
  • Sodium hydroxide- NaOH
  • Potassium hydroxide- KOH
  • Rubidium hydroxide- RbOH
  • Cesium hydroxide- CsOH
  • Calcium hydroxide- Ca(OH)2
  • Strontium hydroxide- Sr(OH)2
  • Barium hydroxide- Ba(OH)2

Calcium, strontium and barium hydroxide can dissociates completely in aqueous solution only when the concentration is 0.01 M or less. But the others in the list dissociates completely, when the concentration of the solution is 1.0 M. There are other strong bases but these are most commonly used.

pH and pOH

The negative logarithm of hydrogen ion concentration is known as the pH of the solution. So, mathematically, we can write that,

pH = -log [H+] …………..(i)

Bases have pH between 7-14. Similarly the negative logarithm of hydroxide ion concentration is known as the pOH of the solution. So,

pOH = -log [OH] ………………(ii)

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Relation between pH and pOH

We know that water is amphoteric. The water dissociates itself to form proton (H+) and hydroxide ion (OH).

H2O ⇋ H+ OH

So, the dissociation constant for Kw for water is,

Kw = [H+] [OH] 

Here the concentration of water is Omitted, as very less amount of water dissociates in this process and the concentration of water remains unchanged or the concentration is constant during this process. Taking logarithm on both sides, we get

log K= log [H+] + log [OH] 

Or -log K= -log [H+] – log [OH] 

So pK= pH + pOH  [from equation (i) and (ii)]

But for pure water,

pKw = 14   [As Kw = 1.0 X 10-14, at 25C]

Thus pH + pOH = 14. ………………..(iii)

Calculation of the pH and pOH of a strong base

The measurement of pH depends on the concentration of hydrogen ion [H+] in aqueous solution. Bases gives hydroxide ions in solution. So, how can we measure the hydrogen ion concentration [H+] of bases in water? We know water dissociates slightly to get proton Hand hydroxide ion OH. It is an equilibrium process. According to Le Chatelier’s principle the equilibrium shift to favorable direction to experience less stress, when any change of condition happened. So, when strong base dissolved in aqueous solution, to reduce the increased concentration of hydroxide ions OH, the equilibrium of the autoionisation of water shifts to left. That means water molecule increases and the concentration of Hion decreases. Thus the pH of the solution of the solution will be different and we can measure that. The pH of this solution will be higher than the neutral or pure water.

Calculation of pOH and pH of strong base (not applied for weak base) can be done in few steps:

Step 1: We need to know the concentration of hydroxide ion [OH]. As strong bases dissociates completely to give hydroxide ions OH. On other way, the same number of moles of strong bases dissociates to give same number of moles of hydroxide ions OH. Thus for strong bases (not applied for weak bases), the concentration of the base and the concentration of the hydroxide ion OH will be same.

Step 2: Calculate the negative logarithm of hydroxide ion [OH] to get the pOH. Because we know from equation (ii), pOH = -log [OH].

Step 3: Calculate the value of pH from the equation (iii), pH + pOH = 14.


Example 1

We will find out the pOH and pH of a 0.10 M NaOH solution.

Step 1: We have already seen that NaOH is a strong base and it completely dissociates in aqueous solution.

NaOH(aq) → Na+(aq) + OH(aq)

So the concentration of hydroxide ion [OH] and the concentration of sodium hydroxide [NaOH] will be same. That means,

[NaOH] = [OH] = 0.10 M

Step 2: The pOH of the solution is

pOH = -log [OH]

⇒ pOH = -log (0.10)

⇒ pOH = 1

Step 3: The pH of the solution is

pH + pOH = 14

⇒ pH + 1 = 14

⇒ pH = 14-1

⇒ pH = 13

Example 2

If we will find out the pOH and pH of a another solution of 0.50 M NaOH then we can compare the result.

Step 1: The concentration of [OH] is

[NaOH] = [OH] = 0.50 M

Step 2: The pOH of the solution is

pOH = -log [OH]

⇒ pOH = -log (0.50)

⇒ pOH = 0.30

Step 3: The pH of the solution is

pH + pOH = 14

⇒ pH + 0.30 = 14

⇒ pH = 14-0.30

⇒ pH = 13.7

If we summarize the result in a table and analyze the changes, we get:

0.500.500.302 X 10-1413.7



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