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Topic: Why wouldn't this work for measuring pH after adding HCl to a buffer.  (Read 1176 times)

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Offline science24

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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.
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 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)
« Last Edit: May 21, 2016, 03:15:20 AM by sjb »

Offline Borek

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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.

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kb=[CA][H+]/[Base]

This is Ka, not Kb.

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So now you add HCl to the mix so the Base + HCl-->CA + H+.

No, that's not a correct reaction equation.

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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.

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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".

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