04-11-2018, 03:13 PM

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)

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)