Question

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

Answer 1

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.