Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: science24 on May 20, 2016, 08:26:48 PM
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A buffer contains 0.4M base and 0.25M conjugate acid. What is the pH when 0.075 moles HCl is added? (pKa of base = 8.68)
Thats the question, the HCl is added to a 1L solution. Also my teacher wrote Ka of Base, I know that's not correct wording, but that is how they worded it. I know the answer is ph=8.68 but why wouldn't it work to do what i did below.
Attempt
I just need an explanation for why my work isnt correct. The initial ph would equal 8.88 since kb=[CA][H+]/[Base] So now you add HCl to the mix so the Base + HCl-->CA + H+. Make an ice table and the base is (.4-x) but the x is negligible so it is just (.4). For the HCl it would be (.075-x) but once again the x is negligible.Then on the right you would have (x) for both CA and H+ since the CA and H+ come from the second reaction so the equation would be Ka=[CA][H+]/[Base][HCl] making it 2.1E-9=(x^2)/(.4)(.075) which you solve for x (or H+) to get x=7.937E-6 then you take the -log(H+) to get the ph of 5.1. Why wouldn't doing this work?
(mod edit remove image and type q as text)
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my teacher wrote Ka of Base, I know that's not correct wording, but that is how they worded it.
Yes, it is a lousy wording - Ka is that of the conjugate acid.
kb=[CA][H+]/[Base]
This is Ka, not Kb.
So now you add HCl to the mix so the Base + HCl-->CA + H+.
No, that's not a correct reaction equation.
Make an ice table and the base is (.4-x) but the x is negligible so it is just (.4)
No, amount of base after the reaction is not 0.4. That was the initial amount, but you added a strong acid.
For the HCl it would be (.075-x)
No, HCl reacted to the end. Besides, you are not interested in the HCl, apparently you are mistaking "conjugate acid" with "any acid present".
Compare http://www.chembuddy.com/?left=buffers&right=pH-change