Question

Question #acfee

Answer 1

See the explanation.

Explanation:

[Part1] Determine the mass of diethylamine.
The reaction between diethylamine and hydrochloric acid is
`(C_2H_5)_2NH + HCl -> (C_2H_5)_2NH_2^+Cl^-`.

Let `m` (g) the mass of diethylamine.
Acording to the reaction formula, the amount of substance is the same for the acid and the base.
`m/73.14 * 1= 0.075 * 15.90/1000 * 1`.
`m=8.72xx10^-2` g.

[Part2] Then, what is the pH at the equivalent point?
The amount of substance of diethylamine chloride(`(C_2H_5)_2NH_2^+Cl^-`) after the titration is
`0.075*15.90/1000 = 1,193xx10^-3` moles, and the molarity is
`(1.193xx10^-3)/((100.00+15.90)/1000)=1.029xx10^-2` M.

The equilibrium state of diethylamine is
`(C_2H_5)_2NH + H_2O ⇄ (C_2H_5)_2NH_2^+ + OH^-`. [1]
The hydrosis of `(C_2H_5)_2NH_2^+` is
`(C_2H_5)_2NH_2^+-` `⇄ (C_2H_5)_2NH + H^+` [2]

And the equilibrium constant is
`K_b=([(C_2H_5)_2NH_2^+][OH^-])/([(C_2H_5)_2NH])=3.09xx10^-4`. [3]

Then, let's solve [3].

[a] `[(C_2H_5)_2NH_2^+]=1.029xx10^-2` M(since `(C_2H_5)_2NH_2^+Cl^-` is a salt)
[b] Let `[H^+]= x` M. Then, the equation [2] tells us that
`[(C_2H_5)_2NH]=x`.
[c] `[OH^-]=K_w/([H^+]) = (1.00xx10^-14)/x` M.

Substitute [a],[b] and [c] to [3].
`(1.029xx10^-2*1.00xx10^-14)/x^2=3.09xx10^-4`
`x^2=(1.029xx10^-2*1.00xx10^-14)/(3.09xx10^-4)=3.330xx10^-13`
`x= 5.770 xx10^-7` M `=[H^+]`.

Therefore, the pH is `-log[H^+]=-log(5.770xx10^-7)~=6.24`.

Answer 2

We've effectively "neutralized" a weak base. Unlike traditional neutralization reactions, the pH at the equivalence point isn't equal to `7`. An `"ICE"` table is useful, here,

Note, I've assumed volumes are additive.

`(C_2H_5)_2NH(aq) + H^(+)(aq) rightleftharpoons (C_2H_5)_2NH_2^(+)(aq)`

are the concentrations at the equivalence point. Now, let's examine another equilibrium that is occurring due to the conjugate acid dissociating:

`(C_2H_5)_2NH_2^(+)(aq) rightleftharpoons H^(+)(aq) + (C_2H_5)_2NH(aq)`

We know the concentration of the conjugate acid, and using the equilibrium expression,

`K_a = ([H^+][(C_2H_5)_2NH])/([(C_2H_5)_2NH_2^(+)]) = (1.0*10^-14)/K_b approx 3.24*10^-11`

Hence,

`3.24*10^-11 = x^2/(0.0120M)`
`x = [H^+] approx 6.23*10^-7M`
`therefore pH = -log[H^+] approx 6.21`

This is reasonable, because the conjugate acid will make the solution slightly acidic at the equivalence point, the weak base having been consumed.