A 25.0 mL sample of 0.100 M propanoic acid (HC3H5O2, Ka = 1.3 ✕ 10-5) is titrated with 0.100 M KOH solution. Calculate the pH after the addition of the following amounts of KOH.
a) 20.0 mL
b) 25.0 mL
c) 30.0 mL
Let's call propanoic acid something simple like HPr to make typing simple.Then
millimols HPr = mL x M = 25.00 x 0.1 M = 2.500.
millimols KOH added = 20 x 0.1 = 2.0
or 25 x 0.1 = 2.50 or 30 x 0.1 = 3.0.
In order that will be as follows.
.............HPr + KOH ==> KPr + H2O
I.............2.50......0..............0.........0
add...................2.0...............................
C............-2.0.....-2.0...........+2.0..................
E...........0.50.........0..............2.0................
So for the 20 mL portion you have a buffer solution; i.e., you have some of the weak acid (the proponoic acid) + the salt. So use the Henderson-Hasslebalch equation.
pH = pKa + log (base)/(acid). The base is the salt. The acid is HPr.
For the 25 mL you can construct an ICE chart as I've done above and you can see that the HPr exactly equals the KOH so that is the equivalence point. Here then you have just the salt; i.e., just KPr. The pH is determined by the hydrolysis of the salt like this.The concn of the salt is millimoles/mL = 2.50/50 = 0.05 M
..................Pr^- + HOH ==> HPr + OH^-
I................0.05......................0...........0
C................-x.........................x...........x
E..................0.05-x.................x...........x
Kb for Pr^- = (Kw/Ka for HPr) = (x)(x)/(0.05-x)
Substitute and solve for x. x = (OH^-) so convert that to H^+ and calculate pH.
For the 30 mL addition, construct and ICE chart and you will see that all of the HPr has been neutralized and you now have extra KOH added after the equivalence point; therefore, you have just a solution of the salt and the KOH So the (OH^-) = millimoles extra KOH/total mL of solution. Convert to H^+ and pH.
Post your work if you get stuck on any of this.
Well, well, well, let's get titrating! Time to crunch some numbers and find out the pH after adding different amounts of KOH.
a) After adding 20.0 mL of KOH, we need to figure out how much acid is left. So, first we calculate the moles of propanoic acid:
moles of propanoic acid = volume of propanoic acid (in L) * molarity of propanoic acid
= 0.025 L * 0.100 M
= 0.0025 moles
Now, let's determine how many moles of KOH were used in the reaction:
moles of KOH = volume of KOH (in L) * molarity of KOH
= 0.020 L * 0.100 M
= 0.002 moles
Since propanoic acid and KOH react in a 1:1 ratio, the moles of propanoic acid remaining is:
moles of propanoic acid remaining = moles of propanoic acid - moles of KOH
= 0.0025 - 0.002
= 0.0005 moles
Now we use the Henderson-Hasselbalch equation to calculate the pH:
pH = pKa + log([A-]/[HA])
pKa = -log(Ka) = -log(1.3 ✕ 10^-5) = 4.89
[HA] = moles of propanoic acid remaining / volume of solution (in L)
= 0.0005 moles / 0.045 L
= 0.011 M
[A-] = moles of KOH / volume of solution (in L)
= 0.002 moles / 0.045 L
= 0.044 M
pH = 4.89 + log(0.044/0.011)
= 4.89 + 1.17
= 6.06
So, after adding 20.0 mL of KOH, the pH will be 6.06.
b) Hold your horses, because after adding 25.0 mL of KOH, the calculation is much simpler. Why? Well, the 25.0 mL of KOH that was added is an exact match to the initial volume of the propanoic acid. That means they have completely reacted, and the pH will be determined by the resulting salt (potassium propanoate) and any remaining KOH. So, the pH will be determined by the concentration of the salt in solution, my friend. We would need to calculate the concentration of potassium propanoate and use that to find the pH.
c) Now, after adding 30.0 mL of KOH, we have more KOH than the initial volume of the propanoic acid. So, after this point, the excess KOH will determine the pH. You can think of it as a KOH party in the solution, taking control of the pH. So, we need to calculate the moles of KOH that are in excess and then determine the concentration of OH- ions. From there, we can find the pOH and then the pH.
Well, I hope this explanation titrated your curiosity and left you with a pH of laughter!
To calculate the pH after the addition of KOH, we can use the Henderson-Hasselbalch equation, which is given by:
pH = pKa + log([A-]/[HA])
Where:
- pH is the pH of the solution
- pKa is the negative logarithm of the acid dissociation constant
- [A-] is the concentration of the conjugate base (in this case, C3H5O2-)
- [HA] is the concentration of the acid (in this case, HC3H5O2)
First, we need to calculate the initial concentration of the acid. We are given:
- Volume of the acid = 25.0 mL
- Concentration of the acid (HC3H5O2) = 0.100 M
Using the equation:
M1V1 = M2V2
where M1 and V1 are the initial concentration and volume, and M2 and V2 are the final concentration and volume, we can calculate the initial moles of the acid:
moles of acid (HC3H5O2) = (0.100 M) x (25.0 mL / 1000 mL/mL) = 0.00250 moles
a) After the addition of 20.0 mL of KOH:
- Total volume = 25.0 mL + 20.0 mL = 45.0 mL
- Moles of KOH = (0.100 M) x (20.0 mL / 1000 mL/mL) = 0.00200 moles
- Moles of acid remaining = 0.00250 moles - 0.00200 moles = 0.00050 moles
- Concentration of acid remaining = 0.00050 moles / (45.0 mL / 1000 mL/mL) = 0.0111 M
Now, we can calculate the concentration of the conjugate base using the fact that KOH is a strong base and will dissociate fully:
Moles of KOH = (0.100 M) x (20.0 mL / 1000 mL/mL) = 0.00200 moles
Concentration of KOH = 0.00200 moles / (45.0 mL / 1000 mL/mL) = 0.0444 M
Substituting these values into the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
pH = -log(1.3 x 10^-5) + log(0.0444 M / 0.0111 M)
pH = 4.89 + log(4.0)
pH = 4.89 + 0.602
pH = 5.49
Therefore, the pH of the solution after the addition of 20.0 mL of KOH is 5.49.
b) After the addition of 25.0 mL of KOH:
- Total volume = 25.0 mL + 25.0 mL = 50.0 mL
- Moles of KOH = (0.100 M) x (25.0 mL / 1000 mL/mL) = 0.00250 moles
- Moles of acid remaining = 0.00250 moles - 0.00250 moles = 0.00000 moles
- Concentration of acid remaining = 0.00000 moles / (50.0 mL / 1000 mL/mL) = 0.00000 M
Substituting these values into the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
pH = -log(1.3 x 10^-5) + log(0 M / 0.00000 M)
pH = 4.89 + log(0/0)
This is undefined, as you cannot take the logarithm of zero. Therefore, the pH cannot be calculated after the addition of 25.0 mL of KOH.
c) After the addition of 30.0 mL of KOH:
- Total volume = 25.0 mL + 30.0 mL = 55.0 mL
- Moles of KOH = (0.100 M) x (30.0 mL / 1000 mL/mL) = 0.00300 moles
- Moles of acid remaining = 0.00250 moles - 0.00300 moles = -0.00050 moles
- Concentration of acid remaining = -0.00050 moles / (55.0 mL / 1000 mL/mL) = -0.0091 M
Substituting these values into the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
pH = -log(1.3 x 10^-5) + log(-0.0091 M / 0.0000 M)
This is undefined, as you cannot take the logarithm of a negative value. Therefore, the pH cannot be calculated after the addition of 30.0 mL of KOH.
To calculate the pH after the addition of KOH, we need to determine the amount of propanoic acid (HC3H5O2) that has reacted with the added KOH and the resulting concentration of the remaining propanoic acid.
First, let's calculate the number of moles of propanoic acid initially present in the 25.0 mL sample:
Moles of HC3H5O2 = Volume of solution (L) x Concentration of HC3H5O2 (mol/L)
= 0.025 L x 0.100 mol/L
= 0.0025 mol
Now, let's consider the reaction between propanoic acid (HC3H5O2) and KOH (potassium hydroxide):
HC3H5O2 + KOH -> KC3H5O2 (potassium propanoate) + H2O
The stoichiometry of this balanced equation tells us that one mole of HC3H5O2 reacts with one mole of KOH to form one mole of KC3H5O2.
Since the concentration of KOH is 0.100 M and the volume added is given for each case, let's now calculate the moles of KOH added:
a) 20.0 mL of KOH added:
Moles of KOH = Volume of solution (L) x Concentration of KOH (mol/L)
= 0.020 L x 0.100 mol/L
= 0.002 mol
b) 25.0 mL of KOH added:
Moles of KOH = 0.025 L x 0.100 mol/L
= 0.0025 mol
c) 30.0 mL of KOH added:
Moles of KOH = 0.030 L x 0.100 mol/L
= 0.003 mol
Now, let's determine the reaction that occurs between the remaining unreacted propanoic acid and the added potassium hydroxide.
Since the moles of KOH added are equal to or greater than the moles of HC3H5O2 initially present, HC3H5O2 will be completely consumed, and only KC3H5O2 will remain.
The remaining volume of the solution after adding KOH will be the sum of the initial volume and the volume of KOH added.
a) 20.0 mL of KOH added:
Remaining volume = 25.0 mL + 20.0 mL = 45.0 mL = 0.045 L
b) 25.0 mL of KOH added:
Remaining volume = 25.0 mL + 25.0 mL = 50.0 mL = 0.050 L
c) 30.0 mL of KOH added:
Remaining volume = 25.0 mL + 30.0 mL = 55.0 mL = 0.055 L
Since KC3H5O2 is a salt of a weak acid and a strong base, the resulting solution will be basic.
To calculate the concentration of KC3H5O2, we divide the moles of KC3H5O2 by the remaining volume of the solution:
a) 20.0 mL of KOH added:
Concentration of KC3H5O2 = Moles of KC3H5O2 / Remaining volume
= 0.002 mol / 0.045 L
= 0.044 M
b) 25.0 mL of KOH added:
Concentration of KC3H5O2 = 0.0025 mol / 0.050 L
= 0.050 M
c) 30.0 mL of KOH added:
Concentration of KC3H5O2 = 0.003 mol / 0.055 L
= 0.055 M
Finally, to calculate the pH of the resulting solution, we need to take the negative logarithm (pH = -log[H+]) of the concentration of the hydronium ions [H+] derived from the ionization of KC3H5O2.
Since KC3H5O2 is a basic salt, it will undergo hydrolysis to produce hydroxide ions (OH-) which react with water to form hydronium ions (H3O+).
The concentration of hydronium ions, [H+], can be calculated using the Kw expression for water:
Kw = [H+][OH-]
[H+] = Kw / [OH-]
Since KC3H5O2 fully dissolves into KC3H5O2 and K+ and OH- ions, the concentration of hydroxide ions [OH-] will be equal to the concentration of KC3H5O2.
a) 20.0 mL of KOH added:
[H+] = Kw / [OH-] = 1.0 x 10^-14 / 0.044 M
pH = -log[H+]
b) 25.0 mL of KOH added:
[H+] = Kw / [OH-] = 1.0 x 10^-14 / 0.050 M
pH = -log[H+]
c) 30.0 mL of KOH added:
[H+] = Kw / [OH-] = 1.0 x 10^-14 / 0.055 M
pH = -log[H+]
By plugging in the values and performing the calculations, you will obtain the pH values for each case.