Question

In: Chemistry

Calculate the pH after addition of a) 0.00 mL b) 25.00 mL c) 50.00 mL d)...

Calculate the pH after addition of

a) 0.00 mL

b) 25.00 mL

c) 50.00 mL

d) 60.00 mL

of 0.1000 M HCl in the titration of 50.00 mL of 0.1000 M hydrazine. Ka(N2H5+)=1.05×10−8.

Please show all steps thank you!

Solutions

Expert Solution

The trick of these problems is to know where is the equivalence point, and exactly in what point of the curve you are. First let's find out the volume to reach the equivalence point:

MaVa = MbVb

Va = 0.1 * 50 / 0.1 = 50 mL

So, part a) and b) we can actually know that we are before of the end point, and in c) is the pH in the equivalence point, and d) would be the excess of acid reacting:

a) at 0 mL

N2H4 + H2O ------> N2H5+ + OH-   Kb = 1x10-14 / 1.05x10-8 = 9.52x10-7

i. 0.1 0 0

e. 0.1-x x x

9.52x10-7 = x2 / 0.1-x You can assume that 0.1-x = 0.1 because Kb is low:

x = (9.52x10-7 * 0.1)1/2

x = 3.09x10-4 M

pOH = -log(3.09x10-4) = 3.51

pH = 14-3.51 = 10.49

b) adding 25 mL

In this point we are in the half equivalence point, so you can assume the following expression:

pOH = pKb or pH = pKa so:

pKa = -logKa = -log(1.05x10-8) = 7.98 = pH

c) in the equivalence point we know that moles of HCl = moles of hydrazine so:

moles of hydrazine = 0.1 * 0.05 = 0.005 moles

Concentration = 0.005 / 0.1 = 0.05 M

The reaction is:

N2H5+ + H2O ------> N2H4 + H3O+

i. 0.05 0 0

e. 0.05-x x x

1.05x10-8 = x2 / 0.05-x

x = (1.05x10-8 * 0.05)1/2

x = 2.29x10-5 M

pH = -log(2.29x10-5) = 4.64

d) in this part, we already pass the equivalence point, so, all we have is the excess of acid added so:

moles or acid = 0.1 * 0.060 = 0.006

moles of remanent acid = 0.006 - 0.005 = 0.001

Concentration = 0.001 / 0.110 = 0.0091 M

pH = -log(0.0091)

pH = 2.04

Hope this helps.


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