Reactions Between Acids and Bases
Lajoy Tucker
Teacher
Introduction to Reactions between Acids and Bases
Acids and bases react in neutralisation reactions where (from acid) + (from base) →
Titration calculations up to this point allow the determination of the number of moles in a solution of unknown concentration through volumetric analysis.
For example, neutralises ³ of . Calculate the volume of required.
Solution:
Moles of
Equation:
1:1 of to ratio, so moles
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The reaction above where n(acid) = n(base) results in a neutral solution. Where either the acid or base is added in excess, the resultant pH of the final solution can still be calculated.
Calculations
When in doubt, moles!
General approach
1. Write the ionic equation
2. Calculate initial moles of acid and base
3. Identify which species is in excess
4. Determine the number of moles of the excess species remaining
5. Calculate the concentration from the remaining moles
6. Calculate pH using appropriate equation
Acids and Bases Explainer Video
Example 1: Strong Acid – Strong Base (simplified)
10 moles of reacts with 8 moles of . Calculate the pH of the final solution (total volume 1 )
Ionic equation | + | ||||
Initial moles (I) | 10* (excess) | 8 (limiting) | Moles of water not required for future pH calculation | ||
Change in moles (C) | -8 | -8 (limiting moles all used up) | |||
End moles (E) | 2 | 0 |
The strong acid is in excess so the pH is calculated directly from [H+]
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Example 2 – Strong Acid – Strong Base
Calculate the pH of the solution formed when of is added to of .
At , has the value .
Ionic equation | + | ||||
Initial moles (I) | 0.050 x 0.1 = 0.0050 mol | 0.500 x 0.05 = 0.025 mol (excess) | Moles of water not required for future pH calculation | ||
Change in moles (C) | -0.0050 | -0.0050 | |||
End moles (E) | 0 | 0.020 |
The base is in excess, so the pH is calculated from
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Example 3 – Weak Acid – Strong Base
a weak acid HA mixed with KOH.
Note - The weak acid only partially dissociates so the concentration of the acid HA is used as the concentration of H+ ions is unknown.
This type of calculation is commonly assessed in the preparation of acidic buffers.
Ionic equation | + | A- | + | ||||
Initial moles (I) | 1.00 x 0.025 = 0.025 (excess) | 0.0405 x 0.050 = 2.025 x 10-3’ (limiting) | 0 | Moles of water not required for future pH calculation | |||
Change in moles (C) | -2.025 x 10-3 | -2.025 x 10-3 | +2.025 x 10-3 | ||||
End moles (E) | 0.022975 | 0 | 2.025 x 10-3 |
The weak acid is in excess therefore Ka is needed to calculate the pH.
Important to note here that [H+] is NOT equal to [A-] as some A- has been made over the course of the reaction.
The volumes in this expression would cancel out so moles can be substituted directly in.
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Example 4 – Half neutralisation of a weak acid
At , the acid dissociation constant for ethanoic acid has the value .
Determine the pH of the solution formed when of potassium hydroxide are added to of ethanoic acid at .
Let ethanoic acid be
Ionic equation | HA | + | A- | + | |||
Initial moles (I) | 0.100 x 0.02 = 0.002 | 0.100 x 0.01 = 0.001 nOH- = ½ (nH+) | 0 | Moles of water not required for future pH calculation | |||
Change in moles (C) | -0.001 | -0.001 | +0.001 | ||||
End moles (E) | 0.001 | 0 | 0.001 |
If the number of moles of base is exactly half that of the weak acid, the acid has been half neutralised.
As seen above, this results in and therefore
Substituting this into the expression, these concentrations cancel out.
Therefore, at half neutralisation,
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Note - Half-neutralisation become more important when interpreting pH curves.
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Exam tip - Spotting when can save a lot of time by skipping many calculation steps
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