Strong bases dissociate completely to hydroxyl ions and related cations in the water. Strong bases show higher pH values than weak bases (such as aqueous ammonia solution). Sodium hydroxide (NaOH), potassium hydroxide (KOH), barium hydroxide are famous strong bases. From those two strong bases, NaOH and KOH has lot of uses in laboratories and chemical industries.
In this tutorial, we will learn following sections of chemistry of bases (alkali)
Strong bases dissociate completely in water. Some of strong bases are NaOH, KOH, Ba(OH)2. In water, these bases dissociate into metal cation and OH- ion. As an example, NaOH dissociates completely to Na+ ion and OH- ion.
MOH(aq) → M+(aq) + OH-(aq)
NaOH(aq) → Na+(aq) + OH-(aq)
KOH(aq) → K+(aq) + OH-(aq)
Following metal hydroxides dissolve very well in water and give strong alkaline solution.
pH is mainly changed due to change of concentration of the solution. The concentration of the base increases, it results increase of pH.
pH is defined for expressing H3O+ concentration and pOH is defined for expressing OH- concentration.
In calculating pH of NaOH solution, first, we have to calculate the pOH value. We have a relationship between pH and pOH as below.
pKa is a constant to a certain temperature and pKa = 14 at 250C. So at 250C,
We can use two methods to calculate pH of basic solution.
Ammonium hydroxide is a weak base give less hydroxyl ion concentration. But sodium hydroxide is a strong acid and completely dissociate to sodium ion and hydroxyl ions.
When hydroxyl ion concentration is high, its basic strength is higher and has a high pH value.
Method 1
The formula of pOH,
calculating pOH is a logarithmic one.
Then use the expression of
to calculate pH. The sum of pH + pOH become 14 when room temperature is 250C. In other temperatures, we have to give the related value of that temperature.
Method 2
There is a relation between [H3O]+(aq) and [OH]-(aq) in aqueous solution.
Kw = constant
At 250Kw,
Kw = 1.0 * 10-14mol2 dm-6
When we know [OH]-(aq), using this equation, we can easily calculate the [H3O]+(aq).
Then, we can use pH equation, to calculate pH.
NOTE
Example 1
The room temperature is 250C.
Note : M = mol dm-3
We solve this example according to method 1.
NaOH dissociate completely into Na+(aq) & OH-(aq) ions in water.
We have to calculate concentration of OH-(aq) to calculate pH.
First we calculate pOH and then calculate the pH using the relationship of pH and pOH
pOH = -log(OH-(aq))
pOH = -log ( 0.1)
pOH = 1
pH + pOH = 14 (at 250C)
pH + 1 = 14
pH = 13
Let's solve example 1 according to method 2.
We know the concentration of OH- ions
[OH-(aq)] = 0.1 moldm-3
Use the relation of,
at 250C, Kw = 1.0 * 10-14mol2 dm-6
[H3O]+(aq) * 0.1 = 1.0 * 10-14
[H3O]+(aq) = 1.0 * 10-13
Now, we know the concentration of [H3O]+(aq), therefore we can use pH equation,
pH = -log10(1.0 * 10-13)
pH = 13
Example 2
Here, we are given the pH value. We have to calculate the concentration of KOH. If we know the pH value, we can calculate pOH from pH + pOH = 14 at 250C.
So pOH = 4
From pOH equation, we can calculate OH- concentration.
Example 3
Here only changes the concentration of NaOH. We studied, NaOH completely dissociate in the water. Therefore,
[OH]-(aq) = 0.01 moldm-3
Then substitute [OH]-(aq) to the pOH equation.
pOH = -log(OH-(aq))
pOH = -log ( 0.01)
pOH = 2
pH + pOH = 14 (at 250C)
pH + 2 = 14
pH = 12
When OH- concentration of aqueous solution is reduced by 10 times, pH is decreased by a 1.
At 250C, pH + pOH = 14
Example 3
Ba(OH)2 is a strong base. Therefore it dissociate completely in water to Ba2+(aq) and OH- ions.
When one Ba(OH)2 molecule dissociate, one Ba2+ ion and two OH- ions are given. See the stoichiometry ratio.
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[OH-(aq)] = 0.2 moldm-3
Then substitute [OH]-(aq) to the pOH equation.
pOH = -log(OH-(aq))
pOH = -log ( 0.2)
pOH = 0.7
pH + pOH = 14 (at 250C)
pH + 0.7 = 14
pH = 13.3
Example 4
4g of solid NaOH are dissolved in water and then diluted the mixture upto 250cm3. Calculate the concentration of OH- ions and pH of solution.
First, number of moles(n) of dissolved NaOH should be calculated.
NaOH = 23 + 16 + 1
NaOH = 40 g mol-1
number of moles of dissolved NaOH, nNaOH
nNaOH = 4 g / 40 gmol-1
nNaOH = 0.1 mol
Calculate the concentration now
[NaOH(aq)] = 0.1mol / 0.25 dm3
[NaOH(aq)] = 0.4 mol dm-3
[OH-(aq)] = 0.4 mol dm-3
pOH = -log(OH-(aq))
pOH = -log ( 0.4)
pOH = 0.4
pH = 13.6
Questions
Yes. it is.
If pH value is high, we know, pH value will completely depend on the concentration of NaOH. Otherwise we have to consider dissociation of water to calculate OH- concentration.
Both NaOH and KOH are strong bases and dissociate completely in the water. So when NaOH and KOH concentrations are equal, after the dissociation, hydroxyl ion concentration is also same. Therefore pH of both solution is equal.
But, basic strength of KOH is much higher than NaOH.
pH of dilute NaOH solutions are usually above 10 if concnetration of NaOH is higher than 0.0001 mol dm-3
pOH is 1 and pH = 13
pOH of 0.01 mol dm-3 NaOH = 2
pH of 0.01 mol dm-3 NaOH = 12
Questions
When sodium hydroxide solution is dissolved in water it dissociates completely to sodium ion and hydroxyl ions. According to the balanced equation, concentraion of hydroxyl ions is equal to the concentration of sodium hydroxide.
Therefore, pH of 0.02 mol dm-3 sodium hydroxide is 1.69
OH- concentration = 0.00092832 mol dm-3. Substitute this in pOH equation. Then you can use pH + pOH = 14 eqution. At 250C, pH + pOH = 14
Nickel hydroxide (Ni(OH)2) is a green color precipitate and give very less OH- amount to water. Therefore, Ni(OH)2 is a weak basic solution. But, KOH is highly soluble in water and dissociates completely in water to give a strong basic solution.
Magnesium hydroxide is a white color precipitate and give very less OH- amount to water. Therefore, OH- concentration is low and show pH values are not higher as NaOH, KOH solutions. According to the given pH, we can calculate what is the concentration of Mg(OH)2.
Assuming all OH- ions exist in water are given by Mg(OH)2 (not from water ionization)
Because, we know the OH- concentration, we can calculate concentration of Mg(OH)2.
Magnesium hydroxide (Mg(OH)2) is not asoluble hydroxide in water. It forms a weak basic solution. First, you should know that, is it possible to have a concentration like 0.32 M of 0.32 M Mg(OH)2. You can calculate that from using solubility data of inorganic compounds.
Yes. KOH is a strong base and it completely dissociates to K+ and OH- ions in the water.
Just like KOH, substitute hydroxyl ion concentration given by NaOH to the pOH equation. Then substitute pOH value to the pH + pOH = 14 equation and find pH of NaOH.
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