calculate the ph for each Calculate the pH for each of the following cases in th

Question

calculate the ph for each Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.140 M HCIO(aq) with 0.140 M KOH (aq). The ionization constant for HCIO can be found here. (a) before addition of any KOH (b) after addition of 25.0 mL of KOH (c) after addition of 35.0 mL of KOH (d) after addition of 50.0 mL of KOH (e) after addition of 60.0 mL of KOH With equal concentrations of monoprotic titrant and analyte, the equivalence point would occur when the added volumes are equal (50.0 mL of KOH added). So at point (c), we are 35.0/50.0 = 70.0% of the way to the equivalence point. If 70.0% of the acid has reacted, then 30.0% remains, and 70.0% of the conjugate base has been formed so [CIO^-]/[HCIO] = 70.0/30.0. Find the pH using the Henderson-Hasselbalch equation pH = pK_a + log [CIO^-]/[HCIO]

Explanation / Answer

Equation:

HClO + KOH KClO + H2O

1 mol HClO reacts with 1 mol KOH to produce 1 mol KClO plus water.

You will notice that throughout most of the problem you have a solution containing a weak acid ( HClO) and the conjugate base of that acid , from the salt KClO. This is a buffer solution .

(a)

Before any addition of KOH

You want to calculate the pH of the acid solution: You do this by using the Ka equation:

Ka of HClO = 4.0 x 10-8

Ka = [H+]2 / [acid]

4.0 x 10-8 = [H+]² / 0.140

[H+]² = (4.0 x 10-8) x 0.140

[H+]² = 5.6 x 10-9

[H+] = (5.6 x 10-9)

[H+] = 7.48 x 10-5 M

So, pH = -log [H+]

      pH = -log (7.48 x 10-5)

     pH = 4.13

(b)

After addition of 25.0 mL of KOH. This produces a mixture of unreacted HClO and KClO in solution. This is a buffer solution. You determine the pH of a buffer solution using the Henderson - Hasselbalch equation.

This particular problem is easily solved - You react 50 mL of 0.140 M HClO with 25 mL of 0.140 M KOH. You have added exactly half the KOH required to neutralize all the acid. You are at the half equivalence point - the concentration of unreacted acid and salt are the same. The pH of this solution is equal to the pKa of the acid:

pKa = - log Ka = -log ( 4.0 x 10-8) = 7.40

Therefore the pH of the buffer is 7.40.

But for completion let us do this using the H-H equation - this will help us with question c) next

Mol HClO in 50.0 mL of 0.140 M HClO solution = 0.050 x 0.140 = 0.007 moles HClO

Mol KOH in 25.0 mL of 0.140 M solution = 0.025 x 0.140= 0.0035 mol KOH

These react to produce 0.0035 mol KClO, and there is 0.0070 - 0.0035 = 0.0035 mol unreacted HClO remaining.

Calculate the molarity of each compound in solution: Final volume = 50+25 = 75 mL = 0.075 L

[HClO] = 0.0035 / 0.075 = 0.047M
[KClO] = 0.0035 / 0.075 = 0.047M

Now apply the H-H equation:

pH = pKa + log ( [salt] /[acid])

pH = 7.40 + log ( 0.047 / 0.047)

pH = 7.40 + log (1.00)

pH = 7.40 + 0.00

pH = 7.40.

(c)

c) After addition of 35 mL of KOH

We will use the above working out, but shortened:

Mol HClO in 50.0 mL of 0.140 M HClO solution = 0.050 x 0.140 = 0.0070 moles HClO

Mol KOH in 40.0ml of 0.22M solution = 0.035 x 0.140 = 0.0049 mol KOH

These react to produce 0.0049 mol KClO, and there is 0.0070 - 0.0049 = 0.0021 mol unreacted HClO remaining.

Calculate the molarity of each compound in solution: Final volume = 50 + 35 = 85 mL = 0.085 L

HClO = 0.0021/0.085 = 0.0247 M
KClO = 0.0049/0.085 = 0.0576 M

Now apply the H-H equation:

pH = pKa + log ( [salt] /[acid])

pH = 7.40 + log ( 0.0576/0.0247)

pH = 7.40 + log 2.33

pH = 7.40 + 0.37

pH = 7.77

(d)

After addition of 50 mL of KOH. What is significant about this titration volume: You are reacting 50 mL of 0.140 M HClO with 50 mL of 0.140M KOH. You are neutralizing all the HClO and producing a solution that contains only KClO. What you are doing in effect is calculating the pH of a KClO solution (no longer a buffer solution - so you do not use the H-H equation)

For our above work , you are reacting 0.0070 mol HClO with 0.0070 mol KOH to produce 0.0070 mol KClO dissolved in 100 mL of solution = 0.1L

Molarity of the KClO solution = 0.0070/0.1 = 0.07 M KClO solution.

Calculate pH of a 0.07 M solution of KClO

KClO dissociates:

KClO K+ + ClO-

The ClO- reacts with water :

ClO- + H2O HClO + OH-

You require the Kb of HClO

Ka of HClO = 4.0 x 10-8

Kb = 10-14 / (Ka)

Kb = 10-14 / (4.0 x 10-8)

Kb = 2.5 x 10-7

Use the Kb equation:

Kb = [HClO] [ OH-] / [KClO]

Because [HClO] = [OH-] , and [KClO] = 0.07 M

Kb = [OH-]2 / 0.07

2.5 x 10-7 = [OH-]2 / 0.07

[OH-]2 = (2.5 x 10-7) x 0.07

[OH-]2 = 1.75 x 10-8

[OH-] = 1.32 x 10-4

To calculate pH you require [H+]

[H+] = 10-14 / [OH-]

[H+] = 10-14 / (1.32 x 10-4)

[H+] = 7.58 x 10-11

pH = -log [H+]

pH = -log (7.58 x 10-11)

pH = 10.12

(e)

After addition of 60 mL of KOH. You saw in question d) above, that 50 mL of KOH had neutralized all the HClO. What you are now doing is adding 10 mL excess of the 0.140 M KOH. The final solution volume is 50 + 60 = 110 mL

Molarity of the KOH solution:

M1V1 = M2V2

M1 x 110 = 0.140 x 10

M1 = 0.013 M KOH solution.

The KClO in solution does not affect the pH of the solution.

Because [KOH] = 0.013 M , then [OH-] = 0.013 M

[H+] = 10-14 / 0.013

[H+] = 7.70 x 10-13 M

pH = -log (7.70 x 10-13)

pH = 12.11

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