Question
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
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
Answers
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.
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.
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