The pH of a 0.164 M aqueous solution of dimethylamine is 11.98. Write the ionization equation,

calculate the values of Kb and pKb and comment on strength of the base....

pH = 11.98

11.98 = -log(H^+)
(H^+) = 1.05E-12

...(CH3)2NH + HOH ==>(CH3)2NH2^+ + OH^-
I....0.164.............0...........0
C.....-x...............x............x
E....0.164-x...........x............x
Since (H^+)(OH^-) = Kw = 1E-14, then
(OH^-)= 1E-14/1.05E-12 = 9.6E-3

Therefore, x = about 9.6E-3 = (OH^-) = [(CH3)2NH].
Substitute those values into Kb expression and solve for Kb, then
pKb = -log(Kb)
Kb = [(CH3)2NH2^+][OH^-]/[(CH3)2NH]

Well, dimethylamine sure knows how to bring the pH up! Let's write the ionization equation first:

(Dimethylamine) + (H2O) <-> (Dimethylammonium) + (OH-)

Now, since the pH is given, we know that the concentration of OH- ions must correspond to a pOH of 14 - 11.98 = 2.02. Using this information, we can calculate the OH- concentration, which is the same as the dimethylammonium ion concentration.

pOH = -log[OH-]
2.02 = -log[OH-]
[OH-] = 10^(-2.02) M

We also know that the concentration of the base (dimethylamine) is 0.164 M. So, we can find the concentration of the unionized dimethylamine by subtracting the concentration of the ionized dimethylammonium from the total concentration:

[Dimethylamine] = [Total] - [Dimethylammonium]
[Dimethylamine] = 0.164 M - 10^(-2.02) M

Now, since Kb is the equilibrium constant for the reaction, we can use the concentrations we just found to calculate it:

Kb = ([Dimethylammonium] * [OH-]) / [Dimethylamine]
Kb = (10^(-2.02) M * 10^(-2.02) M) / (0.164 M - 10^(-2.02) M)

Finally, we can find pKb by taking the negative logarithm of Kb:

pKb = -log(Kb)

And there you have it! You've got all the values you need - Kb and pKb. As for the strength of the base, well, with a pKb close to 14 (since pOH + pKb = 14), dimethylamine is considered a relatively weak base. So, it's not the strongest base at the party, but it still knows how to make the solution a bit more basic!

The ionization equation for dimethylamine (CH3)2NH is:

(CH3)2NH + H2O ⇌ (CH3)2NH2+ + OH-

To calculate the value of Kb, we need the concentration of (CH3)2NH2+ ions. Since dimethylamine is a weak base, we can assume that only a small fraction of it ionizes. At equilibrium, let x represent the concentration of (CH3)2NH2+ ions.

(CH3)2NH + H2O ⇌ (CH3)2NH2+ + OH-
Initial: 0.164 M 0 0
Change: -x +x +x
Equilibrium: 0.164 - x x x

To calculate the value of Kb, we can use the expression:

Kb = [ (CH3)2NH2+ ] [ OH- ] / [ (CH3)2NH ]

We can substitute the equilibrium values into the equation:

Kb = x * x / (0.164 - x)

The pH of the solution is given as 11.98, which means the pOH is 14 - 11.98 = 2.02. Since pH + pOH = 14, the concentration of OH- ions can be calculated as 10^-pOH = 10^-2.02 = 0.00794 M.

Now, we can substitute the concentration of OH- ions into the equation:

7.94 × 10^-3 = x * x / (0.164 - x)

Solving for x using the quadratic formula gives us two possible values, but we only consider the smaller value since it represents the concentration of (CH3)2NH2+ ions:

x = 0.066 M

Finally, we can calculate the value of Kb:

Kb = x * x / (0.164 - x)
= 0.066 * 0.066 / (0.164 - 0.066)
= 0.0049 M

The pKb is calculated using the equation:

pKb = -log10(Kb)
= -log10(0.0049)
= 2.31

Based on the calculated pKb value, we can conclude that dimethylamine is a weak base.

To write the ionization equation for dimethylamine, we need to understand its chemical formula.

Dimethylamine has the formula (CH3)2NH. When it is dissolved in water, it ionizes to produce a hydroxide ion (OH-) and a dimethylammonium ion (CH3)2NH2+.

The ionization equation for dimethylamine can be written as:
(CH3)2NH + H2O -> (CH3)2NH2+ + OH-

Now, let's calculate the value of Kb (the base dissociation constant) and pKb (the negative logarithm of Kb).

To find Kb, we need the concentration values of the dimethylamine and hydroxide ions. In this case, the concentration of dimethylamine is 0.164 M.

Since dimethylamine ionizes into one mole of hydroxide ions, the concentration of OH- is also 0.164 M.

The equation for Kb is:
Kb = [CH3)2NH2+][OH-] / [dimethylamine]

Substituting the concentrations into the equation, we get:
Kb = (0.164)(0.164) / 0.164

Kb = 0.164

Now, let's calculate pKb, which is the negative logarithm of Kb:
pKb = -log(Kb)

Using the value of Kb we found earlier, we get:
pKb = -log(0.164)

Finally, we can comment on the strength of the base.

The strength of a base is determined by its Kb value or pKb value. Generally, a higher value of Kb or a lower value of pKb indicates a stronger base.

In this case, since the value of Kb is 0.164, which is relatively low, we can conclude that dimethylamine is a weak base. Similarly, the value of pKb, which is obtained as a negative logarithm, will also be relatively high for a weak base.

Therefore, based on the calculated Kb value and pKb value, we can say that dimethylamine is a weak base.