Calculate the percent dissociation of the add upon reaching equilibrium in each
ID: 1019195 • Letter: C
Question
Calculate the percent dissociation of the add upon reaching equilibrium in each of the following cases. 0.50 M acetic acid 0.50 M acetic acid 0.0050 M acetic acid You should have observed an increase in the percent dissociation as the concentration deceased from that in to that in. Explain to me using your knowledge of equilibria, why this trend makes sense. Methylamine, CH_3NH_2, has a pK_b of 3.36 and forms hydroxide ions by an indirect pathway when dissolved in water. What is the value of the acid dissociation constant, K_d, for this molecules conjugate? Should methylamine be considered a strong acid, weak acid, strong base, or weak base? If 7.75 g is dissolved in one of water, what will be the equilibrium value for the hydronium ion concentration, pH, hydroxide ion concentration, and pOH?Explanation / Answer
Solution:- (7) Let's take a weak acid as HA. The disosciation of this weak acid could be shown as....
HA(aq) <---------> H+(aq) + A-(aq)
Let's say the concentration of the acid is c and the change in concentration is X. Then the ice table could be made as....
HA(aq) <---------> H+(aq) + A-(aq)
I c 0 0
C -X +X +X
E c - X X X
Let's say the dissociation constant for the acid is Ka. Then the expression will be...
Ka = [H+] [A-]/[HA] = (X)2/(c - X)
If the acid is very weak then its Ka will also be very low, then c - X could be taken as c. So,....
Ka = (X)2/c
X2 = Ka.c
X = square root of (Ka.c)
Percent dissociation = (X/c) * 100
Since, for the percent dissociation X is divided by c, percent dissociation will be more if value of c will decrease.
Let's do the calculations for the given concentrations of acetic acid. Ka for acetic acid is 1.8 x 10-5.
(a) when concentration is 0.50 M.
X = square root of (1.8 x 10-5 x 0.50) = 0.003
percent dissociation = (0.003/0.50) x 100 = 0.6%
(b) when concentration is 0.050 M
X = square root of (1.8 x 10-5 x 0.050) = 0.00095
percent dissociation = (0.00095/0.050) x 100 = 1.9%
(c) when the concentration is 0.0050 M
X = square root of (1.8 x 10-5 x 0.0050) = 0.0003
percent dissociation = (0.0003/0.0050) x 100 = 6%
So, it is also clear from the calculations that percent dissociation increases if the acid is more diluted.
(8) The reaction for the protonation of methylamine could be written as...
CH3NH2(aq) + H2O(l) <---------> , CH3NH3+(aq) + OH-(aq)
The relationship between Ka and Kb is given as...
Ka x Kb = Kw
Ka = Kw/Kb
where Kw is the dissociation constant for water and it's value is 1.0 x 10-14.
Value of pKb is given as 3.36.
we know that, Kb = 10-pKb
Kb = 10-3.36 = 4.37 x 10-4
Now we can easily calculate Ka for the conjugate acid of methylamine.
Ka = 1.0 x 10-14/4.37 x 10-4 = 2.29 x 10-11
Strong and weak acid or base is actually comparble. Given pKb value of methylamine is an indication towards a weak base. Base becomes weaker and weaker as the pKb value start increasing. It's a base since it's Ka is less than Kw. So, it could easily accept proton from water.
In the next part of this problem, 7.75 g of methylamine are dissolved in one liter water. Molar mass of CH3NH2 is 12 + 3(1) + 14 + 2(1) = 12 + 3 + 14 + 2 = 31 g/mol.
moles of methylamine = 7.75 g x (1mol/31g) = 0.25 mol
molarity = 0.25 mol/1L = 0.25 M
Let's make the ice table for the equation we have written above for methylamine.
CH3NH2(aq) + H2O(l) <---------> , CH3NH3+(aq) + OH-(aq)
I 0.25 0 0
C -X +X +X
E 0.25 - X X X
4.37 x 10-4 = (X)2/(0.25 - X)
on cross multiply...
X2 = 4.37 x 10-4(0.25 - X)
X2 = 1.09 x 10-4 - 4.37 x 10-4 X
On rearranging this equation in quadratic equation form...
X2 + 4.37 x 10-4 X - 1.09 x 10-4 = 0
On solving this quadratic equation for X...
X = 0.0102
So, [OH-] = 0.0102 M
pOH = - log[OH-] = - log(0.0102) = 1.99
we know that, pH = 14 - pOH
pH = 14 - 1.99 = 12.01
From pH we can calculate H+ or H3O+(hydronium ion) concentration.
[H3O+] = 10-pH = 10-12.01 = 9.77 x 10-13
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