A 30.00mL sample of 0.150M KOH is titrated with 0.125M HClO4 solution. Calculate the pH after the following volumes of acid have been added: 30.0 mL, 35.0 mL, 36.0 mL, 37.0 mL, 40.0 mL
I am doing similar homeowrk and I was wondering how you calculate the ph after the equivalence point?
This is a strong acid/strong base titration so the pH at the equivalence point will be 7.00. The first thing to do is to calculate where the equivalence point is.The following will do that.
mL acid x M acid = mL base x M base
mL acid x 0.125 = 30.00 x 0.150
mL acid = ?
Then set up an ICE chart for each addition of acid to the base. We start with how many moles of KOH? That is M x L = 0.150 x 0.0300 = .0045
Additions of HClO4 are
30.00 x 0.125 = 0.00375
35.00 --- you can fill
36.00 --- you can fill
40.00 --- you can fill.
..........KOH + HClO4 ==>KClO4 + H2O
initial..0.0045..0.........0......0
add 30..........0.00375............
change.-0.00375 -0.00375..+0.00375..etc
equil....0.00075..0...0.00375..0.00375
The ICE chart tells you what you have at this point in the titration which for 30.00 mL is a solution of KOH and KClO4 in a volume of 60.00 mL (30.00 acid _ 30.00 base).
M KOH = moles/L = 0.00075/0.0600 = 0.0125M KOH. Then pOH = ? and obtain pH from pH + pOH = pKw = 14.
Everything up to equilvanece point is done this way. The equivalence point, as I pointed out above has a pH of 7.00, and everything after the equivalence point done the same way EXCEPT you should note that for those points its the HClO4 in excess. Post your work if you get stuck.
To calculate the pH after different volumes of acid have been added, we need to consider the reaction between KOH and HClO4 and calculate the concentration of OH- ions and H+ ions at each stage.
The balanced equation for the reaction between KOH and HClO4 is as follows:
KOH + HClO4 -> KClO4 + H2O
Since KOH is a strong base and HClO4 is a strong acid, they will react completely, forming water and salt. This means that the concentration of KOH will decrease and the concentration of HClO4 will increase as the acid is added.
Step 1: Initial conditions (before any acid is added)
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Volume of KOH solution (V1) = 30.00 mL
Concentration of KOH (C1) = 0.150 M
To calculate the initial concentration of OH- ions ([OH-]1), we can use the formula:
C1V1 = C2V2
Where C2 is the concentration of OH- ions and V2 is the total volume of the solution after the acid is added.
Using the equation above, we can calculate [OH-]1 as follows:
0.150 M * 30.00 mL = C2 * (30.00 mL + 30.00 mL)
C2 = (0.150 M * 30.00 mL) / (60.00 mL)
C2 = 0.075 M
To calculate the pOH, we can use the formula:
pOH = -log10([OH-])
pOH1 = -log10(0.075) ≈ 1.12
To calculate the pH, we can use the formula:
pH + pOH = 14
pH1 = 14 - pOH1 ≈ 12.88
Step 2: After adding 5.0 mL of HClO4 solution
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Volume of HClO4 solution added = 5.0 mL
Final volume of the solution (V2) = 30.00 mL + 5.00 mL = 35.00 mL
Concentration of HClO4 (C3) = 0.125 M
We can calculate the concentration of H+ ions ([H+]) after adding 5.0 mL of HClO4 solution by using the formula:
C1V1 = C3V3
Where V3 is the volume of HClO4 solution added and [H+] is the concentration of H+ ions.
Using the equation above, we can calculate [H+] as follows:
0.150 M * 30.00 mL = C3 * (5.00 mL)
C3 = (0.150 M * 30.00 mL) / (5.00 mL)
C3 = 0.900 M
To calculate the pH, we can use the formula:
pH = -log10([H+])
pH2 = -log10(0.900) ≈ 0.05
Step 3: After adding 6.0 mL of HClO4 solution
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Volume of HClO4 solution added = 6.0 mL
Final volume of the solution (V4) = 30.00 mL + 6.00 mL = 36.00 mL
Concentration of HClO4 (C5) = 0.125 M
Using the same formula as in Step 2, we can calculate [H+] as follows:
0.150 M * 30.00 mL = C5 * (6.00 mL)
C5 = (0.150 M * 30.00 mL) / (6.00 mL)
C5 = 0.750 M
pH3 = -log10(0.750) ≈ 0.12
Step 4: After adding 7.0 mL of HClO4 solution
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Volume of HClO4 solution added = 7.0 mL
Final volume of the solution (V5) = 30.00 mL + 7.00 mL = 37.00 mL
Concentration of HClO4 (C7) = 0.125 M
Using the same formula as in Step 2, we can calculate [H+] as follows:
0.150 M * 30.00 mL = C7 * (7.00 mL)
C7 = (0.150 M * 30.00 mL) / (7.00 mL)
C7 = 0.643 M
pH4 = -log10(0.643) ≈ 0.19
Step 5: After adding 10.0 mL of HClO4 solution
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Volume of HClO4 solution added = 10.0 mL
Final volume of the solution (V6) = 30.00 mL + 10.0 mL = 40.00 mL
Concentration of HClO4 (C9) = 0.125 M
Using the same formula as in Step 2, we can calculate [H+] as follows:
0.150 M * 30.00 mL = C9 * (10.0 mL)
C9 = (0.150 M * 30.00 mL) / (10.0 mL)
C9 = 0.450 M
pH5 = -log10(0.450) ≈ 0.35
So, the pH after the following volumes of acid have been added are approximately:
30.0 mL: pH ≈ 12.88
35.0 mL: pH ≈ 0.05
36.0 mL: pH ≈ 0.12
37.0 mL: pH ≈ 0.19
40.0 mL: pH ≈ 0.35
To calculate the pH after the given volumes of acid have been added, we need to understand the concept of titration and its relation to acid-base reactions.
In a titration, a known volume and concentration of one solution (the titrant) is added to another solution until a chemical reaction between the two is complete. The point at which the reaction is complete is called the equivalence point.
In this particular titration, we are adding a solution of HClO4 to a solution of KOH. The reaction between HClO4 and KOH is an acid-base reaction, where HClO4 is the acid (donor of H+ ions) and KOH is the base (acceptor of H+ ions).
The balanced equation for this reaction is:
KOH + HClO4 → KClO4 + H2O
From the equation, we can see that one mole of KOH reacts with one mole of HClO4, producing one mole of KClO4 and one mole of water.
To calculate the pH after each volume of acid has been added, we need to determine the number of moles of KOH initially present in the solution, as well as the number of moles of HClO4 that have been added.
1. Calculate the number of moles of KOH initially present:
Moles of KOH = volume of KOH solution (in L) × concentration of KOH (in mol/L)
Volume of KOH solution = 30.00 mL = 0.03000 L (convert mL to L)
Concentration of KOH = 0.150 M
Moles of KOH = 0.03000 L × 0.150 mol/L
2. Determine the limiting reactant:
The limiting reactant is the one that will be completely used up first in the reaction. In this case, whichever reactant we have less moles of will be the limiting reactant. The other reactant will be in excess.
3. Calculate the number of moles of HClO4 added:
In the first case, no HClO4 has been added yet, so the number of moles of HClO4 added is 0.
4. Calculate the number of moles of excess HClO4 after each addition:
Moles of excess HClO4 = moles of HClO4 added - moles of KOH reacted
Since no HClO4 has been added yet, the moles of excess HClO4 is equal to 0 in the first case.
5. Determine the concentration of KOH and HClO4 after each addition:
Concentration of KOH = (moles of KOH remaining) / (total volume of solution)
Concentration of HClO4 = (moles of HClO4 remaining) / (total volume of solution)
Total volume of solution = initial volume of KOH solution + volume of HClO4 added
6. Calculate the moles of OH- and H3O+ ions remaining after each addition:
Moles of OH- remaining = (moles of KOH remaining) - (moles of HClO4 reacted)
Moles of H3O+ formed = (moles of HClO4 reacted) - (moles of OH- reacted)
7. Calculate the pOH and pH after each addition:
pOH = -log10 (concentration of OH-)
pH = 14 - pOH
Using these steps, you can calculate the pH after each volume of acid has been added: 30.0 mL, 35.0 mL, 36.0 mL, 37.0 mL, and 40.0 mL.