I was given a Chemistry question on the titration of a strong base being added to a weak acid. The question was:
"Calculate the pH in a titration when 10.0cm^3 of a 0.10moldm^-3 solution of NaOH is added to a 10.0cm^3 of 0.25moldm^-3 solution of ethanioc acid? (Ka = 1.76 x 10^-5)
I'm completely lost on how to do this. I can work it out for strong acid + strong base (vise versa) but i'm not very good working with Ka. Please can you explain everything in full as i'm not amazing at working out what to put where and in which equation.
Thanks!
The trick to these problem is to recognize what you have in the solution. Let's call ethanoic acid Het.
Then Het + NaOH ==> Naet + HOH
So all of the NaOH is used, some of the Het remains unreacted and some Naet is formed. This is a buffered solution of Het/Naet (a weak acid and its salt).
How much NaOH did we start with?
0.1 mol/dm^3 x 0.01 dm^3 = 0.001 mols NaOH.
How much Het? That is 0.25 mol/dm^3 x 0.01 = 0.0025 mols Het.
Looking at the equation, you can see that all of the NaOH is used but we have some Het remaining. How much is that.
0.0025 mol - 0.001 mol = 0.0015 mols Het remaining.
How much product (Naet) was formed? Of course that is 0.001 mol Naet.
All of that is in a total volume of 10+10 = 20 cc = 0.020 dm^3 (or 0.020 liters).
So final concn of Naet = (0.001/0.020) = ??
Final concn of Het = 0.0015/0.020
(Check these numbers--I get bleary eyed when reading them.)
Now use the Henderson-Hasselbalch equation to solve.
pH = pKa + log (base/acid) = ??
The base is (et) which we calculated above and the acid (Het) we have above also. Just plug and chug.
Aaaaah, thank you DrBob222!
A lot clearer than my chemistry teacher explained it!
Shall work my way through it and see what I can make of it.
Cheers for such a fast response :)
Sure, I'd be happy to explain how to solve this titration problem involving a weak acid and a strong base.
First, let's break down the problem into smaller steps:
Step 1: Write the balanced chemical equation for the reaction between the weak acid (ethanoic acid, CH3COOH) and the strong base (sodium hydroxide, NaOH).
CH3COOH + NaOH -> CH3COONa + H2O
Step 2: Identify the conjugate acid-base pairs in the reaction.
The weak acid (CH3COOH) and its conjugate base (CH3COO-) form one conjugate acid-base pair.
The strong base (NaOH) and its conjugate acid (H2O) form the other conjugate acid-base pair.
Step 3: Calculate the initial moles of the weak acid (ethanoic acid) and the strong base (sodium hydroxide) used in the titration.
Moles of ethanoic acid:
moles = volume (in dm^3) × concentration (in mol/dm^3)
moles = 10.0 cm^3 × 0.25 mol/dm^3 = 0.0025 mol
Moles of sodium hydroxide:
moles = volume (in dm^3) × concentration (in mol/dm^3)
moles = 10.0 cm^3 × 0.10 mol/dm^3 = 0.0010 mol
Step 4: Determine which is the limiting reactant.
The limiting reactant is the one that is completely consumed in the reaction. To determine the limiting reactant, compare the moles of the weak acid (0.0025 mol) and the strong base (0.0010 mol).
Since the moles of the weak acid are higher, it is the limiting reactant. This means that the reaction will consume all of the weak acid and leave some excess strong base.
Step 5: Calculate the moles of excess strong base.
The moles of excess strong base can be calculated by subtracting the moles of the limiting reactant (weak acid) from the total moles of the strong base.
Moles of excess strong base = Total moles of strong base - Moles of limiting reactant
= 0.0010 mol - 0.0025 mol (which is negative, indicating that all of the strong base will be consumed in the reaction)
Step 6: Determine the volume of the solution after the reaction.
The total volume of the solution after the reaction will be the sum of the initial volumes of the weak acid and the strong base.
Volume of solution after reaction = Initial volume of weak acid + Initial volume of strong base
= 10.0 cm^3 + 10.0 cm^3 = 20.0 cm^3
Step 7: Calculate the concentration of the weak acid in the final solution after the reaction.
Concentration = Moles / Volume
Concentration = 0.0025 mol / 20.0 cm^3 = 0.125 moldm^-3
Step 8: Calculate the concentration of the conjugate base (CH3COO-) in the final solution after the reaction.
Since the weak acid is completely converted to its conjugate base, the concentration of the conjugate base will be equal to the concentration of the weak acid in the final solution.
Concentration of conjugate base = 0.125 moldm^-3
Step 9: Apply the Henderson-Hasselbalch equation to calculate the pH of the solution.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log10([Conjugate base] / [Weak acid])
In this case, the conjugate base is CH3COO- (0.125 moldm^-3) and the weak acid is CH3COOH (0.125 moldm^-3).
pH = -log10(Ka) + log10([Conjugate base] / [Weak acid])
= -log10(1.76 x 10^-5) + log10(0.125 / 0.125)
= -log10(1.76 x 10^-5) + log10(1)
Using a calculator, you can find the values of -log10(1.76 x 10^-5) and log10(1), and then subtract the first value from the second value to determine the pH of the solution.
I hope this explanation helps you understand how to solve this titration problem involving a weak acid and a strong base. If you have any further questions, please feel free to ask!