The pH of a 0.164 M aqueous solution of dimethylamine is 11.98. Write the ionization equation,
calculate the values of Kb and pKa
and comment on strength of the
(CH3)2NH + H2O = (CH3)2NH2{+} + OH{-}
Kb = [(CH3)2NH2{+}] [OH{-}]/[(CH3)2NH]
At 11.98 the contribution of OH{-} concentration from dissociated water is negligible.
[OH{-}] = 10^(11.98-14.00)
[(CH3)2NH2{+}] = [OH{-}]
[(CH3)2NH] = 0.164M - [OH{-}]
Kb = [OH{-}]^2 / (0.164M - [OH{-}])
= (0.009549{9})^2 / (0.164M - 0.009549{9})
pKa = 14 - ( -log{10}(Kb) )
Well, well, well, it looks like we've got a chemistry question on our hands! Don't worry, I'll do my best to bring some laughter to this equation.
Okay, let's start with the ionization equation for dimethylamine, which is (CH3)2NH. When it ionizes in water, it forms its conjugate acid, (CH3)2NH2+, and hydroxide ions, OH-.
(CH3)2NH + H2O ⇌ (CH3)2NH2+ + OH-
Now, let's move on to finding the values of Kb and pKa.
pKa is a little trickier to find but don't panic, I'll clown around with it. If we know the pH of the solution, we can use the formula pKa + pKb = 14. Since we're given the pH of 11.98, we can subtract that from 14 to find pKa. So pKa = 14 - 11.98 = 2.02.
Bzzzrt! Now let's calculate the value of Kb. Since Kb is the equilibrium constant for the reverse ionization reaction, we can use the equation:
Kw = Ka * Kb
Here, Kw is the autoionization constant of water, which is 1.0 x 10^-14 at 25 degrees Celsius. And Ka is the acid dissociation constant, which we can get by taking the negative logarithm of pKa:
Ka = 10^(-pKa)
So, Kb = Kw / Ka = 1.0 x 10^-14 / 10^(-2.02).
And voila! With a little bit of math, we've got the value of Kb.
Now, coming to the strength of dimethylamine. A higher value of Kb indicates a stronger base, which means dimethylamine is quite the player in the base game.
So, there you have it! I hope my clowning around made this chemistry problem a little more entertaining for you. Just remember, when in doubt, chemical equations and humor go hand in hand.
To write the ionization equation for dimethylamine (CH3)2NH, we can follow the general form for a weak base:
(CH3)2NH + H2O ⇌ (CH3)2NH2+ + OH-
The value of Kb (base dissociation constant) can be calculated using the equation:
Kb = [CH3)2NH2+][OH-] / [CH3)2NH]
Since we know the concentration of dimethylamine, we can substitute it into the equation:
Kb = [CH3)2NH2+][OH-] / (0.164)
To calculate pKa, we need to find the value of Ka (acid dissociation constant) of the conjugate acid, which in this case is (CH3)2NH2+. The relationship between Ka and Kb is given by the equation:
Ka × Kb = Kw (water dissociation constant)
The value of Kw is 1.0 × 10^-14 at 25°C. Rearranging the equation, we get:
Ka = Kw / Kb
Substituting the known value of Kw and the calculated value of Kb, we can find Ka:
Ka = (1.0 × 10^-14) / Kb
Finally, we can calculate pKa by taking the negative logarithm of Ka:
pKa = -log(Ka)
Based on the pH value of the solution (pH = 11.98), we can determine that the dimethylamine is a weak base. This is because the pH is greater than 7, indicating a basic solution.
To write the ionization equation for dimethylamine, we first need to understand its molecular formula. Dimethylamine is usually represented as (CH3)2NH.
The ionization equation for dimethylamine in water can be written as:
(CH3)2NH + H2O ⇌ (CH3)2NH2+ + OH-
In this equation, dimethylamine molecule (CH3)2NH reacts with water (H2O) to form the dimethylammonium ion ((CH3)2NH2+) and hydroxide ion (OH-).
To calculate the values of Kb and pKa, we need to use the equilibrium expression and given pH value.
Kb is the equilibrium constant for the base reaction, which can be calculated using the following equation:
Kb = [BH+][OH-] / [B]
Here, [BH+] represents the concentration of the conjugate acid (dimethylammonium ion), [OH-] represents the concentration of the hydroxide ion, and [B] represents the concentration of the base (dimethylamine).
Given that the pH of the solution is 11.98, we can calculate the concentration of hydroxide ion ([OH-]) using the equation:
[OH-] = 10^(-pOH)
Since pOH is the negative logarithm of the hydroxide ion concentration, we can calculate it as follows:
pOH = 14 - pH = 14 - 11.98 = 2.02
Therefore, [OH-] = 10^(-pOH) = 10^(-2.02)
Next, we can consider the concentration of the base ([B]). The concentration of dimethylamine in the solution is given as 0.164 M, which represents the concentration of the base.
Finally, we need to calculate the concentration of the conjugate acid ([BH+]). Since dimethylamine (CH3)2NH is a weak base, it does not fully ionize. Therefore, we can assume that the concentration of the conjugate acid ([BH+]) is negligible compared to the concentration of the base ([B]).
Using the calculated values of [OH-], [B], and [BH+], we can substitute them into the Kb equation and solve for Kb:
Kb = ([BH+][OH-]) / [B]
Given that the concentration of [BH+] is negligible compared to [B], we can simplify the equation as:
Kb ≈ [OH-] / [B]
Substituting the values of [OH-] and [B], we can calculate Kb.
After calculating Kb, to find pKa, we need to use the relationship between pKa and pKb. In this case, since we are dealing with a base (dimethylamine), we can use the equation:
pKa + pKb = 14
Substituting the calculated value of Kb into this equation, we can solve for pKa.
Once both Kb and pKa values are obtained, you can comment on the strength of the base. If Kb is a large value (greater than 1), it indicates that the base is strong. If Kb is a small value (less than 1), it indicates that the base is weak. Similarly, if pKa is a small value (less than 7), it indicates that the conjugate acid is strong. If pKa is a large value (greater than 7), it indicates that the conjugate acid is weak.