Question

Calculate the pH for each of the cases in the titration of 25.0 mL of 0.180 M pyridine, C 5 H 5 N ( aq ) with 0.180 M HBr ( aq ) . A. before addition of any HBr B. after addition of 12.5 mL of HBr C. after addition of 24.0 mL of HBr D. after addition of 25.0 mL of HBr E. after addition of 34.0 mL of HBr

Answer 1

Answer:

Explanations

Calculate the pH for each of the following cases in thetitration of 25.0 mL of 0.100 M pyridine, C5H5N(aq) with 0.100 MHBr(aq):

a.Before and addition of HBr

b.After addition of 12.5ml HBr

c.After addition of 15ml HBr

d.After addition of 25ml HBr

e.After addition of 33ml HBr

SOLUTION ;;;

Kb of pyridine =1.5*10^-9

a)let the dissociation be x.so,

kb=x^2/(0.1-x)

or 1.5*10^-9=x^2/(0.1-x)

or x=1.225*10^-5

so,

[OH-]=1.225*10^-5

so,

pOH=-log([OH-])

so pH=9.088

b)now this will effectively behave as a buffer

pKb=8.82

so pOH=pKb+log(salt/acid)

=8.82+log((12.5*0.1)/(25*0.1-12.5*0.1))

=8.82

so pH=14-pOH

=5.18

c)again using the same equation as the above,

pOH=pKb+log(salt/acid)

=8.82+log((15*0.1)/(25*0.1-15*0.1))

=9

so pH=14-9

=5

d)now the base is completely neutralised.so,

concentration of the salt formed=0.1/2

=0.05 M

so,

pH=7-0.5pKb-0.5log(C)

=7-0.5*8.82-0.5*log(0.05)

=3.24

e)concentration of H+=(33*0.1-25*0.1)/(33+25)

=0.01379

so pH=-log(0.01379)

=1.86

Answer 2

Answer:

Check Explanation.

Explanation:

(A). Before the addition of any acid at all, meaning that we only have a base. Therefore,

The concentration of the [OH^-] = √(kb × [pyridine].

The concentration of the [OH^-] = [√(1.7 x 10^-9 × 0.180M).

[OH^-] = 1.75 × 10^-5.

Recall that the formula for Calculating pOH is given below;

pOH= - log [OH-].

pOH = - log (1.75 x 10^-5).

pOH = 4.76.

Hence;

pH = 14.0 - pOH .

pH= 14.0 - 4.76.

pH = 9.24.

(B). After the addition of 12.5 mL of HBr. And this means that we should Calculate the pH at midpoint of the titration.

To solve this part we will first find the ka of the buffer acid and then using the value of the ka to find its pka.

(Recall; kb = 1.7 x 10^-9).

==> ka = 1 x 10^-14 / 1.7 x 10^-9.

ka= 5.9 x 10^-6.

==> pKa = - log Ka.

pKa= - log (5.9 x 10^-6).

pKa = 5.23.

So, back to the solution;

pH = pKa + log ([C5H5N] / [C5H5NH+]).

pH= 5.23 + log [0.180]/ [0.180].

pH = 5.23 + log 1.

pH = 5.23.

(C). Then, after addition of 24.0 mL of HBr. Therefore, we have used 24mL out of 25mL and it now remain 1mL.

pH = 5.23 + log (1.0 / 24.0 ).

pH= 5.23 - 1.4.

pH = 3.83.

(D). after addition of 25.0 mL of HBr.

( Recall ka = 5.9 x 10^-6).

So, we will first Calculate the number of moles of C5H5N.

C5H5N = 0.180 × 25.0.

C5H5N = 4.5 mmoles

Therefore, in this stage we are now having 4.5 mmoles of C5H5NH+ specie.

Hence, [C5H5NH+] = 4.5 / 50.0 mL = 0.09M.

Next, we Calculate [H3O+] from;

[H3O+] = √(0.09 × 5.9 x 10^-6).

[H3O+] = 5.31 × 10^-7.

Therefore,

pH = -log [H3O+] .

pH = - log (5.31 x 10^-7).

pH = 6.3.

(E). Then, after addition of 34.0 mL of HBr.

First, we Calculate the number of mmoles of HBr;

0.180 × 34 = 6.12.

Therefore, the excess mmoles= 6.12 - 4.5 = 1.62.

Since, we now have an extra 9mL, then, the concentration of the acid=

1.62 / 59mL.

= 0.027 M.

Then, pH= -log [0.027].

pH= 1.046.