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Effect on pH of Diluting an Aqueous Solution of a Strong Base
Sodium hydroxide, NaOH, is a strong base.
It completely dissociates in water to produce sodium ions, Na+(aq), and hydroxide ions, OH-(aq), as shown in the balanced chemical equation below:
| word equation:
|| sodium hydroxide
|| sodium ions
|| hydroxide ions
| balanced chemical equation:
Let's start by making a stock solution of aqueous sodium hydroxide by dissolving 1 pellet of NaOH(s) with a mass of 0.100 g in enough water to make 100 mL of sodium hydroxide solution, NaOH(aq).
We can calculate the concentration of this NaOH solution, c(NaOH(aq)), in moles per litre (mol L-1 or M) as shown below:
concentration of NaOH(aq) = moles(NaOH dissolved in water) ÷ volume of solution in litres
c(NaOH(aq)) = n(NaOH(aq)) ÷ V(solution)
The amount of NaOH in moles is given by:
moles(NaOH) = mass(NaOH in grams) ÷ molar mass(NaOH in grams per mole)
n(NaOH) = m(NaOH in grams) ÷ M(NaOH in grams per mole)
So the amount of NaOH in moles of 0.100 g of NaOH with a molar mass of 22.99 + 16.00 + 1.008 = 39.998 g mol-1 is:
n(NaOH) = 0.100
g ÷ 39.998 g mol-1 = 2.50 × 10-3 mol
The volume of the solution is 100 mL, which we can convert to litres by dividing by 1000 mL/L:
V(L) = 100
mL ÷ 1000 mL/L = 0.100 L
and the concentration of this solution is:
c(NaOH(aq)) = (2.50 × 10-3 mol) ÷ (0.100 L) = 2.50 × 10-2 mol L-1
From our balanced chemical equation for the complete dissociation of NaOH, we see that the stoichiometric ratio (mole ratio) of NaOH : OH- is 1:1
So, 1 mole of NaOH produces 1 mole of OH-, and 2.50 × 10-3 mol NaOH produces 2.50 × 10-3 mol OH-(aq).
Therefore the concentration of hydroxide ions in our solution must be the same as the theoretical concentration of sodium hydroxide we calculated above:
c(OH-(aq)) = c(NaOH(aq)) = 2.50 × 10-2 mol L-1
For an aqueous solution the relationship between the concentration of hydroxide ions in solution, [OH-(aq)], and the concentration of hydrogen ions in solution, [H+(aq)], is given by the ion-product for water (Kw)
Kw = [OH-(aq)][H+(aq)]
at 25°C and atmospheric pressures, Kw = 10-14 so
10-14 = [OH-(aq)][H+(aq)]
for our aqueous solution of sodium hydroxide at 25°C
10-14 = [2.50 × 10-2][H+(aq)]
and if we divide both sides of the equation by 2.50 × 10-2 we will determine the concentration of hydrogen ions in the solution:
(10-14) ÷ (2.50 × 10-2) = ([
2.50 × 10-2][H+(aq)]) ÷ ( 2.50 × 10-2)
4.00 × 10-13 mol L-1 = [H+(aq)]
We can convert this hydrogen ion concentration to a pH using an equation
pH = -log10[H+(aq)] = -log10[4.00 × 10-13] = 12.40
An alternative route is to calculate the pOH of the solution:
pOH = -log10[OH-(aq)] = -log10[2.50 × 10-2] = 1.60
and then subtract this pOH from the pKw for water (14.00 at 25°c) in order to find the pH of the solution:
pH = 14.00 - pOH = 14.00 - 1.60 = 12.40
We can summarise our findings about this stock solution of sodium hydroxide as shown below:
|| 2.50 × 10-2 mol L-1
|| 4.00 × 10-13 mol L-1
Now let's dilute this stock solution by pouring all 100 mL of our stock solution of NaOH(aq) into a 250 mL volumetric flask and fill that up to the mark with distilled water.
The amount of NaOH dissolved in the water hasn't changed, that is still 2.50 × 10-3 mol.
So, because NaOH completely dissociates water, the amount of hydroxide ions dissolved in the water is also 2.5 × 10-3 mol
But now those ions are suspended in a much greater volume of solution, 250 mL instead of 100 mL.
We can calculate the concentration of this dilute solution, c(dilute):
c(dilute) = moles(OH-) ÷ volume of solution in litres
c(dilute) = (2.50 × 10-3 mol) ÷ (250
mL/1000 mL/L) = 1.00 × 10-2 mol L-1
By adding water to the stock solution we have diluted it, and, the concentration of OH-(aq) has decreased from 2.50 × 10-2 mol L-1 in the stock solution to 1.00 × 10-2 mol L-1 in the dilute solution.
Let's calculate the pOH of the dilute solution:
pOH = -log10[OH-(aq)] = -log10[1.00 × 10-2] = 2.00
And we can use this to calculate the pH of the solution using pKw = 14.00 (at 25°C):
pH = 14.00 - pOH = 14.00 - 2.00 = 12.00
Then we can calculate the concentration of hydrogen ions in the dilute solution:
[H+(aq)] = 10-pH = 10-12 = 1.00 × 10-12 mol L-1
Now we will compare the concentrations of OH-(aq) and H+(aq), as well as the pOH and pH, for the stock solution and the dilute solution as shown in the table below:
|| 2.50 × 10-2 mol L-1
|| 4.00 × 10-13 mol L-1
|| 1.00 × 10-2 mol L-1
|| 1.00 × 10-12 mol L-1
- the concentration of hydroxide ions, [OH-(aq)], is lower in the dilute solution than in the stock solution
- the pOH of dilute solution is greater than the pOH of the stock solution
- the concentration of hydrogen ions, [H+(aq)], is higher in the dilute solution than in the stock solution
- the pH of the dilute solution is less than the pH of the stock solution.
What happens if you keep adding water to this solution?
Could you keep diluting your stock solution until you arrive at a very, very, dilute solution, say a solution with a hydroxide ion concentration of 10-10 mol L-1?
What is the pH of this solution?
pH = 14.00 - pOH
pH = 14.00 - (-log10[OH-(aq)]
pH = 14.00 - (-log1010-10] = 14.00 - 10.00 = 4.00
But this answer doesn't make sense does it?
You can't keep adding water to a basic solution and end up with an acidic solution (pH = 4 at 25°C).
In fact, we have been ignoring the dissociation of water itself which also increases the number of hydroxide ions and hydrogen ions in solution.
As long the concentration of hydroxide ions provided by the base is much greater than 10-7 mol L-1 we can comfortably ignore water's contribution to the hydroxide ion concentration.
But, as soon as the concentration of hydroxide ions provided by the base is much less than 10-7 mol L-1 it is the water that is contributing the majority of the hydroxide ions and we can ignore the contribution made by the base.
In short, as we dilute an aqueous solution of strong base at 25°C, the [OH-(aq)] approaches 10-7 mol L-1, and the pH of the solution approaches 7.00.
How to Calculate the Volume of Water to Add to an Aqueous Solution of Strong Base to Achieve a Specified pH
Sodium hydroxide is a substance that absorbs water and carbon dioxide from the air.
As a consequence it is very difficult to get an accurate or preceise measurement of the mass of sodium hydroxide, and therefore it follows that a solution made using this inaccurate and imprecise mass will have an inaccurately known concentration.
It is usually better to use a stock solution of NaOH(aq) whose concentration has been determined by titration for example, and then dilute that down to achieve the desired concentration or pH.
If we know the concentration and volume of stock solution to be used, and, the desired pH of the dilute solution, we can calculate what the total volume of the dilute solution must be, and, therefore we can determine how much water needs to be added to the stock solution.
Imagine you have been given 25.00 mL of an aqueous solution of 0.0200 mol L-1 NaOH(aq) with which to make a new solution with a pH of 10.00 (at 25°C).
How much water will you have to add to the 25.00 mL of stock solution?
For an aqueous solution at 25°C, pH + pOH = 14.00
So we can calculate the pOH of the diluted solution:
pOH(dilute) = 14.00 - pH(dilute) = 14.00 - 10.00 = 4.00
Now we can calculate the concentration of hydroxide ions in the dilute solution, c(dilute):
c(dilute) = 10-pOH = 10-4 mol L-1
Recall the "dilution equation" : c1V1 = c2V2
c1 = concentration of OH- in stock solution = 0.0200 mol L-1
V1 = volume of stock solution used = 25.00 mL = 25.00
mL ÷ 1000 mL/L = 0.02500 L
c2 = concentration of OH- in dilute solution = c(dilute) = 10-4 mol L-1
V2 = total volume of dilute solution produced = ? L
which we can now use to determine the total volume of the new, dilute solution, V2, in litres:
| Dilution equation :
| Divide both sides of the equation by c2 :
| Substitute the values into the equation:
|mol L-1 × 0.02500 L
mol L-1 | =
| and solve:
|| 5.00 L
We have calculated the total volume of the new, dilute solution to be 5.00 L.
5.00 L = 5.00
L × 1000 mL/ L = 5000 mL
So, how much water must we add to the 25.00 mL of stock solution?
total volume (dilute solution) = volume of stock used + water added
5000 mL = 25.00 mL + water added
5000 mL - 25.00 mL = 4975 mL = water added
Assuming additivity of volumes ( that is, 1 mL of solution + 1 mL water = 2 mL of solution), we need to add 4975 mL of water to 25.00 mL of 0.0200 mol L-1 NaOH(aq) to make a new solution with a pH of 10.00