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Effect on pH of Mixing Aqueous Solutions of Strong Acid with a Strong Base
Imagine you have two beakers; Beaker A and Beaker B, both at 25°C and atmospheric pressure.
Beaker A contains an acid: 100 mL of 0.100 mol L-1 aqueous hydrochloric acid solution with a pH of 1
Beaker B contains a base: 10 mL of 0.01 mol L-1 aqueous sodium hydroxide solution with a pH of 12
These two labelled beakers are shown in the diagram below:
Beaker A
100 mL
[HCl(aq)] = 0.100 mol L-1
pH = 1
|
|
Beaker B
10 mL
[NaOH(aq)] = 0.010 mol L-1
pH = 12
|
Hydrochloric acid is a strong acid, it completely dissociates (or ionises) in water producing hydrogen ions (H+(aq)) and chloride ions (Cl-(aq)) according to the following chemical equation:
HCl(aq) → H+(aq) + Cl-(aq)
Sodium hydroxide is a strong base, it completely dissociates (or ionises) in water producing sodium ions (Na+(aq)) and hydroxide ions (OH-(aq)) as shown in the chemical equation below:
NaOH(aq) → Na+(aq) + OH-(aq)
If we mix the hydrochloric acid in Beaker A with the sodium hydroxide base in Beaker B a neutralisation reaction occurs in which the hydrogen ions (H+(aq)) react with the hydroxide ions (OH-(aq)) to produce water molecules (H2O(l)).
A salt (an ionic compound), sodium chloride, is also formed but this is soluble in water (NaCl(aq)) so it really exists in the solution as mobile sodium ions (Na+(aq)) and mobile chloride ions (Cl-(aq)).
We can use several different chemical equations to represent this neutralisation reaction as shown below:
| molecular equation: |
HCl(aq) |
+ |
NaOH(aq) |
→ |
H2O(l) |
+ |
NaCl(aq) |
|---|
| ionic equation: |
H+(aq) + Cl-(aq) |
+ |
Na+(aq) + OH-(aq) |
→ |
H2O(l) |
+ |
Na+(aq) + Cl-(aq) |
|---|
| net ionic equation: |
H+(aq) |
+ |
OH-(aq) |
→ |
H2O(l) |
|
|
|---|
After we mix the contents of these two beakers together, by pouring the contents of both beakers into a third beaker (Beaker R) for example, what will be the pH of the final solution (the pH of the resultant solution)?
Beaker A
100 mL
[HCl(aq)] = 0.100 mol L-1
pH = 1
|
|
Beaker B
10 mL
[NaOH(aq)] = 0.010 mol L-1
pH = 12
|
| |
↓ |
|
↓ |
|
| |
Beaker R
pH = ?
|
|
Is the final pH of the resultant solution in Beaker R:
- pH = 6.5 (half-way between pH 1 and pH and 12) ?
But remember that pH is a logarithmic scale!
Therefore the final pH will NOT be half-way between 1 and 12.
- pH = 7 ?
But has all the acid neutralised all the base?
n(H+(aq)) = n(OH-(aq)) ?
- pH > 7 ?
But is there an excess of base ?
n(OH-(aq)) > n(H+(aq)) ?
And how will we calculate the pH ?
- pH < 7 ?
But is there an excess of acid ?
n(H+(aq)) > n(OH-(aq)) ?
And how will we calculate the pH ?
Clearly the first thing we need to do is calculate the moles of hydrochloric acid in Beaker A and the moles of sodium hydroxide in Beaker B.
To do this we will use the molarity equation as shown below:
n = c × V
n = moles of solute (either n(HCl(aq)) or n(NaOH(aq)))
c = concentration of solution in mol L-1 or M (either c(HCl(aq)) or c(NaOH(aq)))
V = volume of solution in L (either V(HCl(aq)) or V(NaOH(aq)))
Remember to convert volumes in mL to L!
V(mL) ÷ 1000 mL/L = V(L)
We have performed these calculations for Beaker A (acid) and Beaker B (base) as shown below:
Beaker A : pH = 1
V(HCl(aq)) = 100 mL = 100 mL/1000 mL/L = 0.100 L
c(HCl(aq)) = 0.100 mol L-1
n(HCl(aq)) = c(HCl(aq)) × V(HCl(aq))
n(HCl(aq)) = 0.100 × 0.100 = 0.0100 mol
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|
Beaker B : pH = 12
V(NaOH(aq)) = 10 mL = 10 mL/1000 mL/L = 0.010 L
c(NaOH(aq)) = 0.010 mol L-1
n(NaOH(aq)) = c(NaOH(aq)) × V(NaOH(aq))
n(NaOH(aq)) = 0.010 × 0.010 = 0.000100 mol
|
Recall that HCl(aq) is a strong acid, it fully dissociates (ionises) in water.
| Dissociation equation |
HCl(aq) |
→ |
H+(aq) |
+ |
Cl-(aq) |
The stoichiometric ratio (mole ratio) of HCl(aq) to H+(aq) to Cl-(aq) is 1:1:1
So, n(HCl(aq)) = n(H+(aq)) = n(Cl-(aq)) = 0.0100 mol
Recall that NaOH(aq) is a strong base, it fully dissociates (ionises) in water.
| Dissociation equation |
NaOH(aq) |
→ |
Na+(aq) |
+ |
OH-(aq) |
The stoichiometric ratio (mole ratio) of NaOH(aq) to Na+(aq) to OH-(aq) is 1:1:1
So, n(NaOH(aq)) = n(Na+(aq)) = n(OH-(aq)) = 0.000100 mol
When we mix the contents of Beaker A (acid) with the contents of Beaker B (base) in Beaker R (resultant solution) then there are 0.0100 moles of H+(aq) available to react and 0.000100 moles of OH-(aq) available to react:
| |
Beaker R : pH = ?
| net ionic equation: |
H+(aq) |
+ |
OH-(aq) |
→ |
H2O(l) |
| available moles: |
0.0100 mol |
+ |
0.000100 mol |
→ |
? |
|
|
Using the stoichiometric ratio (mole ratio) for the neutralisation reaction given by the net ionic equation above, we see that the ratio of H+(aq) to OH-(aq) to H2O(l) is 1:1:1
1 mole of H+(aq) reacts with 1 mole of OH-(aq) to produce 1 mol of H2O(l)
0.0100 mol of H+(aq) reacts with 0.0100 mol of OH-(aq) to produce 0.0100 mol of H2O(l)
0.000100 mol of H+(aq) reacts with 0.000100 mol of OH-(aq) to produce 0.000100 mol of H2O(l)
Look at the moles of H+(aq) and OH-(aq) available to react in Beaker R above.
Can you see that there are more moles of H+(aq) available to react than there are moles of OH-(aq)
| n(H+(aq)) |
> |
n(OH-(aq)) |
| 0.0100 mol |
> |
0.000100 mol |
H+(aq) is the reactant in excess.
OH-(aq) is the limiting reagent.
Therefore, of the available 0.0100 mol of H+(aq), only 0.000100 mol WILL react with ALL of the available OH-(aq).
This means that there will be some moles of H+(aq) remaining in solution, some H+(aq) will not have reacted.
We can calculate the moles of H+(aq) that will remain in solution (the excess H+(aq)):
n(H+(aq)(excess)) = n(H+(aq)(available)) - n(H+(aq)(reacted))
n(H+(aq)(excess)) = 0.0100 mol - 0.000100 mol
n(H+(aq)(excess)) = 0.0099 mol
So, after mixing the acid in Beaker A with the base in Beaker B the resultant solution in Beaker R will contain 0.000100 moles of extra water as a result of the neutralisation reaction and 0.0099 moles of excess H+(aq)!
Yes, it will also contain 0.000100 moles of NaCl(aq), but because this is soluble in water and doesn't react with any of the reactants or products we are ignoring it.
We can represent the events occurring as a result of the neutralisation reaction in Beaker R after the acid and base have been mixed as shown in the diagram below:
| |
Beaker R : pH = ?
| net ionic equation: |
H+(aq) |
+ |
OH-(aq) |
→ |
H2O(l) |
| available moles: |
0.0100 mol |
+ |
0.000100 mol |
→ |
? |
| reacting moles: |
0.000100 mol |
|
0.000100 mol |
|
|
| moles after reaction |
0.0099 mol |
|
0 mol |
|
0.000100 mol |
|
|
In order to calculate the pH of the resultant solution in Beaker R above we need to know the concentration of hydrogen ions in mol L-1, [H+(aq)] or c(H+(aq)).
To calculate the concentration of H+(aq) in solution we will use the equation given below:
c(H+(aq)) = n(H+(aq)) ÷ V(solution in L)
We have already calculated the moles of H+(aq) above, n(H+(aq)) = 0.0099 mol
And we can calculate the volume of the solution in Beaker R by assuming additivity of volumes, that is if you add 1 mL of acid solution to 1 mL of base solution the result will be 1 mL + 1 mL = 2 mL of solution.
V(solution) = volume of HCl(aq) + volume of NaOH(aq) + volume of water produced by neutralisation
Because concentration will be expressed in units of moles per litre (mol L-1) we want the volumes of each reagent to be units of litres (L).
You may recall that we performed this calculation at the very beginning of this section when we introduced Beaker A and Beaker B.
volume of HCl(aq) = V(HCl(aq)) = 0.100 L
volume of NaOH(aq) = V(NaOH(aq)) = 0.0100 L
volume of H2O(l) produced = ? L
What is the volume of 0.000100 moles of H2O(l) ?
First we can calculate the mass of H2O(l) produced by the reaction:
mass (g) = moles (mol) × molar mass (g mol-1)
mass(H2O(l)) = moles((H2O(l)) × molar mass(H2O(l))
mass(H2O(l)) = 0.000100 × (2 × 1.008 + 16.00)
mass(H2O(l)) = 0.000100 × 18.016
mass(H2O(l)) = 0.0018016 g
Then we can recall that the density of water at 25°C is 1.00 g mL-1
So, 1 g of water occupies a volume of 1 mL
0.0018016 g of water occupies a volume of 0.0018016 × 1 mL = 0.0018016 mL
Convert this volume of water in mL to a volume in L by dividing the volume in mL be 1000:
0.0018016 mL = 0.0018016 mL ÷ 1000 mL/L = 0.0000018016 L = 1.8016 × 10-6 L
The volume of water produced by the reaction is so small compared to the volume of the HCl(aq) and the volume of the NaOH(aq) mixed together to make the resultant solution in Beaker R that it can reasonably be said to be negligible and we will ignore it.
So the volume of solution in Beaker R (after mixing the HCl(aq) and NaOH(aq) together) is:
V(solution) = volume of HCl(aq) + volume of NaOH(aq)
V(solution) = 0.100 L + 0.0100 L
V(solution) = 0.110 L
And now we can calculate the concentration of hydrogen ions in the resultant solution in Beaker R using the equation:
c(H+(aq)(excess)) = n(H+(aq)(excess)) ÷ V(solution)
n(H+(aq)(excess)) = 0.0099 mol
V(solution) = 0.110 L
c(H+(aq)(excess)) = 0.0099 mol ÷ 0.110 L
c(H+(aq)(excess)) = [H+(aq)(excess)] = 0.090 mol L-1
Now we can calculate the pH of the resultant solution in Beaker R (after the acid and base were mixed and allowed to react) using the equation given below:
pH = -log10[H+(aq)]
We calculated the concentration of hydrogen ions remaining in solution in Beaker R after the acid and base were mixed:
[H+(aq)(excess)] = 0.090 mol L-1
But what about all the water that is present as the solvent in the solution?
Won't the self-dissociation of water increase the concentration of H+(aq) in the resultant solution in Beaker R?
Kw = [H+(aq)(water)][OH-(aq)(water)] = 10-14 (25°C, 1 atm)
water is neutral so [H+(aq)(water)] = [OH-(aq)(water)] so we can write:
Kw = [H+(aq)(water)]2
[H+(aq)(water)]2 = 10-14
take the square root of both sides of the equation:
√[H+(aq)(water)]2 = √10-14
[H+(aq)(water)] = 10-7 mol L-1 = 0.0000001 mol L-1
Can you see that the concentration of hydrogen ions produced by the neutralisation reaction, [H+(aq)(excess)], is much, much larger than the concentration of hydrogen ions produced by the self-dissocation of water, [H+(aq)(water)] ?
[H+(aq)(excess)] >> [H+(aq)(water)]
0.090 mol L-1 >> 0.0000001 mol L-1
So we will ignore the concentration of hydrogen ions produced by the self-dissociation of water and assume all the hydrogen ions in the resultant solution are due to the excess HCl(aq) added to the NaOH(aq):
[H+(aq)] = [H+(aq)(excess)] = 0.090 mol L-1
And now we can calculate the pH of the resultant solution in Beaker R after mixing the acid in Beaker A with the base in Beaker B:
pH = -log10[H+(aq)]
pH = -log10[0.090]
pH = 1.05
We can represent this whole process in the diagram given below:
Beaker A : pH = 1
V(acid) = 100 mL
[HCl(aq)] = 0.100 mol L-1
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|
Beaker B : pH = 12
V(base) = 10 mL
[NaOH(aq)] = 0.010 mol L-1
|
| |
↓ |
|
↓ |
|
| |
Beaker R : pH = 1.05
V(solution) = 110 mL
[H+(aq)] = 0.090 mol L-1
|
|
When 100 mL of 0.100 mol L-1 HCl(aq) with a pH of 1 is mixed with 10 mL of 0.010 mol L-1 NaOH(aq) with a pH of 12, the resultant solution has a volume of 110 mL with a hydrogen ion concentration of 0.090 mol L-1 and a pH of 1.05.
There is one more important question left to ask about the resultant solution in Beaker R.
Since all the sodium hydroxide solution (NaOH(aq)) was neutralised by the hydrochloric acid (HCl(aq)) does that mean there are no hydroxide ions (OH-(aq)) in the resultant solution in Beaker R?
What is the concentration of hydroxide ions in the resultant solution in Beaker R?
All the NaOH(aq) has been neutralised by the HCl(aq) which left 0.0099 mol of H+(aq) remaining in the resultant solution with a concentration of 0.090 mol L-1.
BUT, there is an awful lot of water in this resultant solution, about 110 mL, which is about 110 g, which is about 6 moles of water!
So the self-dissocation of water (self-ionisation of water) plays an important role in determining the concentration of hydroxide ions in the resultant solution (OH-(aq)) because this is effectively the ONLY source of OH-(aq) in the resultant solution in Beaker R.
Kw = [H+(aq)][OH+(aq)] = 10-14 (at 25°C, 1 atm)
We know the concentration of hydrogen ions in the solution because we calculated that above:
[H+(aq)] = 0.090 mol L-1
We can use this to calculate the concentration of OH-(aq) in the resultant solution in Beaker R:
| [H+(aq)][OH-(aq)] |
= |
10-14 |
|
| 0.090 × [OH-(aq)] |
= |
10-14 |
|
0.090 × [OH-(aq)]
0.090 |
= |
10-14
0.090 |
|
| [OH-(aq)] |
= |
10-14
0.090 |
|
| [OH-(aq)] |
= |
1.11 × 10-13 mol L-1 |
Although the concentration of hydroxide ions in the resultant solution is very, very, small, it is not negligible because the self-dissociation of water is the ONLY source of hydroxide ions in the solution.
And we can, ofcourse, use this hydroxide ion concentration to calculate the pOH of the resultant solution in Beaker R:
pOH(resultant solution) = -log10[OH-(aq)]
pOH(resultant solution) = -log10[1.11 × 10-13]
pOH(resultant solution) = 12.95
Ofcourse, we could have taken a short-cut if we remembered that, for an aqueous solution at 25°C and 1 atm:
pH + pOH = 14
So, since we know that the pH of the resultant solution in Beaker R is 1.05 we can substitute that into the equation and solve for pOH:
pH + pOH = 14
1.05 + pOH = 14
1.05 - 1.05 + pOH = 14 - 1.05
pOH = 12.95
