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1) What is the pH at the equivalence point in the titration of a 28.1 mL sample

ID: 1042821 • Letter: 1

Question

1) What is the pH at the equivalence point in the titration of a 28.1 mL sample of a 0.424 M aqueous hypochlorous acid solution with a 0.469 M aqueous barium hydroxide solution?

2)A 26.1 mL sample of 0.261 M diethylamine, (C2H5)2NH, is titrated with 0.361 M perchloric acid.

The pH before the addition of any perchloric acid is .  

Use the Tables link in the References for any equilibrium constants that are required.

3)A 21.6 mL sample of 0.207 M trimethylamine, (CH3)3N, is titrated with 0.321 M hydroiodic acid.

After adding 20.2 mL of hydroiodic acid, the pH is .

Use the Tables link in the References for any equilibrium constants that are required.

Explanation / Answer

1)
Ka of hypochlorous acid = 3.0*10^-8

1 mol of Ba(OH)2 has 2 mol of OH-
So,
[OH-] = 2*[Ba(OH)2]
= 2*0.469 M
= 0.938 M

find the volume of OH- used to reach equivalence point
M(HClO)*V(HClO) =M(OH-)*V(OH-)
0.424 M *28.1 mL = 0.938M *V(OH-)
V(OH-) = 12.7019 mL

Given:
M(HClO) = 0.424 M
V(HClO) = 28.1 mL
M(OH-) = 0.938 M
V(OH-) = 12.7019 mL


mol(HClO) = M(HClO) * V(HClO)
mol(HClO) = 0.424 M * 28.1 mL = 11.9144 mmol

mol(OH-) = M(OH-) * V(OH-)
mol(OH-) = 0.938 M * 12.7019 mL = 11.9144 mmol


We have:
mol(HClO) = 11.9144 mmol
mol(OH-) = 11.9144 mmol

11.9144 mmol of both will react to form ClO- and H2O

ClO- here is strong base
ClO- formed = 11.9144 mmol
Volume of Solution = 28.1 + 12.7019 = 40.8019 mL
Kb of ClO- = Kw/Ka = 1*10^-14/3*10^-8 = 3.333*10^-7
concentration ofClO-,c = 11.9144 mmol/40.8019 mL = 0.292M

ClO- dissociates as

ClO-        + H2O   ----->     HClO +   OH-
0.292                        0         0
0.292-x                      x         x


Kb = [HClO][OH-]/[ClO-]
Kb = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes

Kb = x*x/(c)
so, x = sqrt (Kb*c)
x = sqrt ((3.333*10^-7)*0.292) = 3.12*10^-4

since c is much greater than x, our assumption is correct
so, x = 3.12*10^-4 M



[OH-] = x = 3.12*10^-4 M

use:
pOH = -log [OH-]
= -log (3.12*10^-4)
= 3.5059


use:
PH = 14 - pOH
= 14 - 3.5059
= 10.4941
Answer: 10.49

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