Calculate the pH of the solution that results from mixing the following reagents. Refer to table 1.
mix 0.189 L of 0.019 M (CH3)2NH(aq) with 0.327 L of 0.011 M (CH3)2NH2Cl(aq)
I tried using the Henderson-Hasselbach equation by finding the pKa but it didn't work....I am still unsure of how to go about setting up the Kb (or Ka) expression in problems like this. Kb is 5.4E-4 and pKb is 3.27
Thanks
HH works very well with this. Dimethyl amine is the base and dimethyl amine hydrochloride is the acid. Calculate the moles of each from M x L and plug in. Use Ka ONLY if you use the HH equation. This is EXACTLY like the HBrO and NaBrO I worked earlier (if you want to use the Kb expression BUT these are bases and salts and not acids and salts). I repeat, however, if you use the H-H equation, use Ka. That is Kw/Kb.
I'm confused how I would use the Ka equation in the HH equation??? I thought HH was just pH=pka + log([base]/[acid]) ...
It is.
moles base = L x M = 0.189 x 0.019 = ??
moles acid = L x M = 0.327 x 0.011 = ??
Ka = Kw/Kb and you know Kb and Kw. You can convert to pKa.
I see you know pKb so 14-pKb = pKa which is the easy way if you know pKb is correct.
To calculate the pH of the solution, we need to determine whether the resulting solution is acidic, basic, or neutral. In this case, we are mixing (CH3)2NH (dimethylamine) with (CH3)2NH2Cl (dimethylammonium chloride). These substances are a weak base and its conjugate acid, respectively.
To find the pH, we need to calculate the concentration of the base and the acid in the resulting solution. Let's start by determining the moles of each substance:
Moles of (CH3)2NH = concentration × volume = 0.019 M × 0.189 L = 0.003591 moles
Moles of (CH3)2NH2Cl = concentration × volume = 0.011 M × 0.327 L = 0.003597 moles
Since (CH3)2NH and (CH3)2NH2Cl react to form a buffer solution, we need to compare the moles of the base and the acid. We can see that they are almost equal, indicating that the resulting solution will be close to neutral.
At this point, we can use the Kb value (base dissociation constant) to calculate the concentration of hydroxide ions ([OH-]) in the solution. The Kb value represents the equilibrium constant for the reaction of the base with water, forming hydroxide ions and the conjugate acid.
The Kb expression for (CH3)2NH is:
Kb = [OH-][CH3)2NH]/[(CH3)2NH2]
We know the value of Kb is 5.4E-4, and the concentration of (CH3)2NH is 0.003591 moles. The concentration of (CH3)2NH2Cl can be considered negligible since its moles are very similar to the moles of (CH3)2NH.
Now we can rearrange the equation to find the concentration of OH-:
[OH-] = (Kb × [(CH3)2NH]) / [(CH3)2NH2]
[OH-] = (5.4E-4 × 0.003591) / 0.003597
[OH-] = 5.394E-7 M
Since the solution is a buffered solution with equal concentrations of the base and the acid, we can assume that the concentration of [H+] (hydronium ion) is equal to the [OH-]. This assumption is valid because the Kb and Ka values are very close, indicating that the solution is close to neutral.
Now, we can use the equation for the pH:
pH = -log[H+]
Since [H+] = [OH-] = 5.394E-7 M, we can calculate the pH:
pH = -log(5.394E-7) = 6.27
Therefore, the pH of the resulting solution is approximately 6.27.