Calculate the ph after the addition of 0.00,5.00,15.00,25.00,40.00,49.00,50.00,5

Question

Calculate the ph after the addition of 0.00,5.00,15.00,25.00,40.00,49.00,50.00,51.00,55.00,and 60.00ml of reagent in the titration of 50.0ml of A. 0.01000M chloroacetic acid with 0.01000M NaOH B. 0.01000M anilinium chloride with 0.1000M NaOH C. 0.1000M hypochlorite acid with 0.1000M NAOH Calculate the ph after the addition of 0.00,5.00,15.00,25.00,40.00,49.00,50.00,51.00,55.00,and 60.00ml of reagent in the titration of 50.0ml of A. 0.01000M chloroacetic acid with 0.01000M NaOH B. 0.01000M anilinium chloride with 0.1000M NaOH C. 0.1000M hypochlorite acid with 0.1000M NAOH A. 0.01000M chloroacetic acid with 0.01000M NaOH B. 0.01000M anilinium chloride with 0.1000M NaOH C. 0.1000M hypochlorite acid with 0.1000M NAOH

Explanation / Answer

In order for you to do this, you will need the pKa or Ka of the acids. I will show you how to do it with the first acid:
Ka Chloroacetic = 1.4x10-3 --> pKa = 2.85

Now, the key to solve this, is to know at which point of the titration you are basing on the equivalence point.

In this case:
0.01 * 50 / 0.01
V of NaOH = 50 mL.

at Vb = 0.
The reaction is as follow:
r: CH2ClCOOH <----------> H+ + CH2ClCOO-
i: 0.01 0 0
e: 0.01-x x x

1.4x10-3 = x2 / 0.01-x --> But Ka is small, and the value of x will be small too, so we can neglect the substract:
1.4x10-3 * 0.01 = x2
x = [H+] = 0.00374 M
pH = -log(0.00374) = 2.43

At 5 mL of base added:
In this case, the moles of base begins to react with the acid, so let's calculate the moles remaining of acid:
moles Acid = 0.01 * 0.05 = 0.0005 moles
moles Base = 0.01 * 0.005 = 0.00005 moles

CH2ClCOOH + OH- <----------> H2O + CH2ClCOO-
0.0005 0.00005 0 0
0.0005-0.00005 0.00045

Now, we can calculate by the following expression the pH:
pH = pKa + log(A-/HA)
pH = 2.85 + log(0.00045/0.0005)
pH = 2.8

At 15 mL of base added, use the same procedure as before to calculate the pH (just change the moles of NaOH and the moles produced because of this change).

At 25 mL of base added. At this point, we use half of the moles of acid, and this half used is the half that it will produced, so the ratio A-/HA = 1. And the pH = pKa. In this case, pH = 2.85

You will do the same procedure at 40 and 49 mL. Now at 50 mL (The equivalence point), the moles of acid are all consumed and the moles of base begins to be in excess, so the form of the acid that is predominant here is CH2ClCOO-, and so the reaction taking place would be:

r: CH2ClCOO- + H2O <-----------> CH2ClCOOH +OH-   Kb =1x10-14 / 1.4x10-3= 7.14x10-12
i: 0.0005 0 0
e: 0.0005-x x x

7.14x10-12 = x2 / 0.0005-x
7.14x10-12 * 0.0005 = x2
x = [OH-] = 5.98x10-6 M
pOH = -log(5.98x10-6) = 7.22
pH = 14 - 7.22
pH = 6.78

From here on, the predominant moles and the excess would be the moles of OH-, so at 55 mL:
moles of OH- = 0.01 * 0.055 = 0.00055 moles
moles of acid = 0.0005
moles of OH- remaining = 0.00005 moles

Now, as the reaction is as above, the expression to use would be:
pOH = pKb + log(B-/BH) ---> pKb = 14 - pKa = 14 - 2.85 = 11.15
pOH = 11.15 + log(0.00005 / 0.00055)
pOH = 10.11
pH = 14 - 10.11 = 3.89

Now, use this procedure to get the pH at 60 mL and then the same with the others.

Hope this helps

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