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A chemist titrates 110.0 mL of a 0.7684 M methylamine (CH3NH2) sotion with 0.4469 M HNO3 solution at 25 °C. Calculate the pH at equivalence. The p K, of methylamine is 3.36 Round your answer to 2 decimal places Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of HNO3 solution added

User Omerkudat
by
8.3k points

1 Answer

2 votes

Answer: The pH at equivalence point for the given solution is 5.59.

Step-by-step explanation:

At the equivalence point,


n_{HNO_(3)} = n_{CH_(3)NH_(2)}

So, first we will calculate the moles of
CH_(3)NH_(2) as follows.


n_{CH_(3)NH_(2)} = 0.764 M * (110 ml)/(1000 ml/L)

= 0.0845 mol

Now, volume of
HNO_(3) present will be calculated as follows.

Volume =
\frac{\text{no. of moles}}{\text{Molarity}}

=
(0.0845)/(0.4469 M)

= 0.1891 L

Therefore, the total volume will be the sum of the given volumes as follows.

110 ml + 189.1 ml

= 299.13 ml

or, = 0.2991 L

Now,
[CH_(3)NH_(3)^(+)] = (0.0845 mol)/(0.2991 L)

= 0.283 M

Chemical equation for this reaction is as follows.


CH_(3)NH_(3)^(+) + H_(2)O \rightleftharpoons CH_(3)NH_(2) + H_(3)O^(+)

As,
k_(a) = (k_(w))/(k_(b))

=
(10^(-14))/(10^(-3.36))

=
2.29 * 10^(-11)

Now,
[HNO_(3)] = \sqrt{k_(a)[CH_(3)NH_(3)^(+)]}

=
\sqrt{2.29 * 10^(-11) * 0.283}

=
2.546 * 10^(-6)

Now, pH will be calculated as follows.

pH =
-log [H_(3)O^(+)]

=
-log (2.546 * 10^(-6))

= 5.59

Thus, we can conclude that pH at equivalence point for the given solution is 5.59.

User Santo Boldizar
by
7.8k points
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