Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.150 M HClO(aq) with 0.150 M KOH(aq). The ionization constant for HClO is 4.0*10^-8
a)before addition of any KOH
b)after addition of 25.0 mL of KOH
c)after addition of 40.0 mL of KOH
d)after addition of 50.0 mL of KOH
e)after addition of 60.0 mL of KOH
a) The HClO is the only significant source of H+ ions initially, so the concentration of H+ ions can be calculated by using the concentration of HClO and the ionization constant for HClO:
[H+] = √(Kw * [HClO]) = √(1.0 * 10^-14 * 0.150) = √(1.5 * 10^-15) = 1.22 * 10^-8 M
Using the formula for pH: pH = -log[H+], we can calculate the pH:
pH = -log(1.22 * 10^-8) ≈ 7.91
b) After adding 25.0 mL of KOH, we can calculate the moles of HClO remaining:
moles of HClO = (initial concentration of HClO) * (initial volume of HClO - volume of KOH added) = 0.150 M * (50.0 mL - 25.0 mL) = 0.150 M * 25.0 mL = 0.00375 moles
The moles of KOH added can be calculated using the concentration and volume of KOH added:
moles of KOH = (concentration of KOH) * (volume of KOH added) = 0.150 M * 25.0 mL = 0.00375 moles
Since KOH is a strong base, it completely reacts with HClO in a 1:1 ratio, producing the same number of moles of OH- ions. We can calculate the concentration of OH- ions:
[OH-] = (moles of OH-) / (total volume) = (0.00375 moles) / (50.0 mL + 25.0 mL) = (0.00375 moles) / (75.0 mL) = 0.00375 moles / 0.075 L = 0.05 M
To calculate the concentration of H+ ions, we can use the ionization constant for water and the concentration of OH- ions:
[H+] = Kw / [OH-] = 1.0 * 10^-14 / 0.05 M = 2.0 * 10^-13 M
Using the formula for pH: pH = -log[H+], we can calculate the pH:
pH = -log(2.0 * 10^-13) ≈ 12.70
c) Same as (b), we use the same method with the new volumes and calculate:
[OH-] = (moles of OH-) / (total volume) = (0.00375 moles) / (50.0 mL + 40.0 mL) = (0.00375 moles) / (90.0 mL) = 0.00375 moles / 0.090 L = 0.0417 M
[H+] = Kw / [OH-] = 1.0 * 10^-14 / 0.0417 M = 2.4 * 10^-13 M
pH = -log(2.4 * 10^-13) ≈ 12.62
d) After adding 50.0 mL of KOH, all the HClO has reacted with KOH and the solution will contain only OH- ions. Since the concentration of OH- ions is the same as the concentration of KOH, we can calculate it directly:
[OH-] = (concentration of KOH) = 0.150 M
[H+] = Kw / [OH-] = 1.0 * 10^-14 / 0.150 M = 6.67 * 10^-14 M
pH = -log(6.67 * 10^-14) ≈ 13.18
e) After adding 60.0 mL of KOH, there will be excess KOH in the solution. The concentration of OH- ions can be calculated using the concentration and volume of KOH added:
[OH-] = (moles of KOH) / (total volume) = (0.00375 moles) / (50.0 mL + 60.0 mL) = (0.00375 moles) / (110.0 mL) = 0.00375 moles / 0.110 L = 0.034 M
[H+] = Kw / [OH-] = 1.0 * 10^-14 / 0.034 M = 2.94 * 10^-13 M
pH = -log(2.94 * 10^-13) ≈ 12.53
So, the pH for each case is:
a) 7.91
b) 12.70
c) 12.62
d) 13.18
e) 12.53
To determine the pH at different points in the titration, we need to consider the reaction between HClO and KOH. The balanced equation is:
HClO + KOH -> KClO + H2O
The initial concentration of HClO is 0.150 M, and the initial volume is 50.0 mL. Since the volumes of HClO and KOH are the same, we can assume that the concentration of KOH is also 0.150 M.
Let's go step by step to calculate the pH at each point in the titration:
a) Before addition of any KOH:
At this point, no KOH is added, so the solution only contains HClO. The pH can be calculated using the equation for weak acid:
pH = -log[H+]
To get [H+], we need to find the concentration of HClO that has ionized. The ionization constant for HClO is 4.0 * 10^-8. Since the initial concentration of HClO is 0.150 M, we can assume that a small fraction of it has ionized based on the ionization constant.
[H+] = √(Ka * [HClO])
[H+] = √((4.0 * 10^-8) * (0.150))
[H+] = √(6.0 * 10^-9)
[H+] ≈ 2.45 * 10^-4 M
pH = -log(2.45 * 10^-4) ≈ 3.61
So, the pH before the addition of any KOH is approximately 3.61.
b) After addition of 25.0 mL of KOH:
At this point, 25.0 mL of 0.150 M KOH is added. The balanced equation indicates that a 1:1 ratio takes place, so we need to calculate the concentration after mixing.
Initial moles of HClO = (0.150 M) * (50.0 mL) = 7.50 mmol
Moles of KOH added = (0.150 M) * (25.0 mL) = 3.75 mmol
Since the reaction is 1:1, the final moles of HClO after the reaction will be (7.50 mmol - 3.75 mmol) = 3.75 mmol.
The final volume after mixing is (50.0 mL + 25.0 mL) = 75.0 mL.
[HClO] = (3.75 mmol) / (75.0 mL) = 0.050 M
Now we can calculate the [H+] and pH using the same method as in part a):
[H+] = √(Ka * [HClO])
[H+] = √((4.0 * 10^-8) * (0.050))
[H+] = √(2.0 * 10^-9)
[H+] ≈ 4.47 * 10^-5 M
pH = -log(4.47 * 10^-5) ≈ 4.35
So, the pH after the addition of 25.0 mL of KOH is approximately 4.35.
c) After addition of 40.0 mL of KOH:
Following the same calculations as in part b), we find that the remaining moles of HClO will be (7.50 mmol - 6.00 mmol) = 1.50 mmol.
The final volume after mixing is (50.0 mL + 40.0 mL) = 90.0 mL.
[HClO] = (1.50 mmol) / (90.0 mL) = 0.0167 M
[H+] = √(Ka * [HClO])
[H+] = √((4.0 * 10^-8) * (0.0167))
[H+] = √(6.68 * 10^-10)
[H+] ≈ 8.18 * 10^-6 M
pH = -log(8.18 * 10^-6) ≈ 5.09
So, the pH after the addition of 40.0 mL of KOH is approximately 5.09.
d) After addition of 50.0 mL of KOH:
At this point, all of the HClO will be neutralized by KOH.
The final volume after mixing is (50.0 mL + 50.0 mL) = 100.0 mL.
[HClO] = 0 M (neutralized)
[H+] = √(Ka * [HClO])
[H+] = √((4.0 * 10^-8) * (0))
[H+] = √(0)
[H+] = 0 M
pH = -log(0) (undefined)
So, the pH after the addition of 50.0 mL of KOH is undefined.
e) After addition of 60.0 mL of KOH:
At this point, all of the HClO will be neutralized by KOH, and an excess of KOH will be present.
The final volume after mixing is (50.0 mL + 60.0 mL) = 110.0 mL.
[HClO] = 0 M (neutralized)
[H+] = √(Ka * [HClO])
[H+] = √((4.0 * 10^-8) * (0))
[H+] = √(0)
[H+] = 0 M
pH = -log(0) (undefined)
So, the pH after the addition of 60.0 mL of KOH is also undefined.
To calculate the pH during the titration of HClO with KOH, you need to consider the reaction between the two and the resulting concentrations of the reactants and products. Here's how to calculate the pH for each case:
a) Before addition of any KOH:
Since no KOH has been added, the solution only contains 0.150 M HClO. To calculate the pH, use the dissociation of HClO:
HClO(aq) ⇌ H+(aq) + ClO-(aq)
Since HClO is a weak acid, we can assume that the concentration of HClO remaining undissociated is still 0.150 M. Therefore, the concentration of H+ ions (which determines the pH) is also 0.150 M. pH = -log[H+] = -log(0.150) = 0.82.
b) After addition of 25.0 mL of KOH:
At this point, you have to determine how much HClO and KOH remain and what new compounds are formed. To do this, we need to calculate the number of moles of each compound involved.
The number of moles of HClO initially present can be calculated using the equation: moles = concentration × volume.
moles of HClO = 0.150 M × 50.0 mL = 0.0075 mol HClO
Since the stoichiometry of the reaction is 1:1 between HClO and KOH, the number of moles of KOH added is equal to the number of moles of HClO consumed.
moles of KOH added = number of moles of HClO consumed = 0.0075 mol KOH
Now, let's calculate the new concentrations of HClO and OH- ions:
For HClO:
The new volume of the solution after adding KOH is 50.0 mL + 25.0 mL = 75.0 mL. Therefore,
new concentration of HClO = moles of HClO / new volume = 0.0075 mol / 75.0 mL = 0.100 M HClO
For OH-:
The concentration of OH- ions can be calculated using the equation: concentration = moles / volume.
concentration of OH- = moles of KOH added / new volume = 0.0075 mol / 75.0 mL = 0.100 M OH-
Since KOH is a strong base, it completely ionizes in water to produce OH- ions. Therefore, the concentration of OH- ions is equal to the concentration of KOH added.
Now, we can calculate the concentration of H+ ions using the ionization constant for HClO.
[H+][ClO-] / [HClO] = 4.0 × 10^-8
Since the ClO- concentration is equal to the H+ concentration (stoichiometry of the reaction), we can use the following equation:
[H+]^2 / [HClO] = 4.0 × 10^-8
Substituting the values, we get:
[H+]^2 / 0.100 = 4.0 × 10^-8
To solve for [H+], rearrange the equation:
[H+]^2 = 4.0 × 10^-8 × 0.100
Taking the square root of both sides:
[H+] = √(4.0 × 10^-8 × 0.100)
Finally, calculate the pH using the equation: pH = -log[H+].
c) After addition of 40.0 mL of KOH:
Follow the same method as in part b) using the new volume and adjusting the concentration of HClO and OH- accordingly.
d) After addition of 50.0 mL of KOH:
Follow the same method as in part b) using the new volume and adjusting the concentration of HClO and OH- accordingly.
e) After addition of 60.0 mL of KOH:
Follow the same method as in part b) using the new volume and adjusting the concentration of HClO and OH- accordingly.
The secret to these problems is to identify what you have in solution. That will tell you how to solve the problem. This is the titration of a weak acid with a strong bsse.
a. The beginning point is just a soln of 0.150M HA (in this case HClO). I can show you how to do that if you don't remember how but I expect you know how to do it.
HA ==> H^+ A^-
0.150...0...0
-x......x...x
0.150-x..x...x
Then substitute into the Ka expression solve for H^+ and convert to pH.
c. 50.00 mL obviously is the equialence point so this is done by the hydrolysis of the salt.
............A^- + HOH ==> HA + OH^-
I........0.0750M............0....0
C...........-x..............x....x
E........0.0750-x............x....x
Kb for the A^- = (Kw/Ka for HA) = (x)(x)/(0.0750-x) and solve for x = OH^-, then convert to pH.
Where did I get the 0.075M for the salt. That's 1/2 initial concn HA and when you titrate equal molarities of monoprotic acid and base it always works out to be 1/2 the initial.
b. All point between a and c are buffer problems.
d. All points after c are just excess OH.
I shall be happy to check your numbers or explain further.