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.
The pH in each case depends on the relative amounts of acid and base present.
a) Before addition of HBr;
Number of moles of C5H5N = 25/1000 L × 0.180 M = 0.0045 moles
pOH = -log[0.0045 M]
pOH = 2.35
pH = 14 - 2.35 = 11.65
b) After addition of 12.5 mL of HBr;
Since the reaction is 1:1
Number of moles of HBr = 12.5/1000 L × 0.180 M = 0.00225 moles
The HBr is the limiting reactant
Amount of excess acid = 0.0045 M - 0.00225 = 0.00225 moles
Total volume of solution = 25mL + 12.5 mL = 37.5mL or 0.0375 L
molarity of excess H+ = 0.00225 moles / 0.0375 L = 0.06 M
pH = - log[0.06 M]
pH = 1.22
c) After addition of 24.0 mL of HBr;
Number of moles of acid = 24/1000 × 0.180 M = 0.00432 moles
Number of moles of excess acid = 0.0045 moles - 0.00432 moles = 0.00018 moles
Total volume of solution = 24.0 mL + 25.0 mL = 49mL or 0.049 L
Molarity of excess acid = 0.00018 moles/ 0.049 L = 0.00367 M
pH = -log[0.00367 M] = 2.4
d) After addition of 25.0 mL of HBr
Number of moles of acid = 25/1000 × 0.180 M = 0.0045 moles
We have an equal number of acid and base hence pH = 11.65
e) After addition of 34.0 mL of HBr;
Number of moles of acid = 34/1000 × 0.180 M =0.00612 moles
We can see that the base is the limiting reactant.
Number of moles of excess base = 0.00162 moles
Total volume of solution = 34 mL + 25 mL = 59mL or 0.059 L
Molarity of excess base = 0.00162 moles/0.059 L = 0.027 M
pOH = -log[ 0.027 M]
pOH = 1.56
pH = 14 - 1.56 = 12.44
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