You're on the right track; you're using the wrong signals.
...........HA + OH^- ==> A^- + H2O
initial..0.153..0.........0......0
added...........x..................
change.....-x...-x........x.......x
equil...0.153-x..0.........x
4.18 = 4.33 + log(x/153-x)
Solve for x. I get about 0.06 but you need to do it more accurately. That makes
acid left = 0.153-0.06 and base = 0.06.
Check it by substituting base and acid numbers into the HH equation and make sure you get 4.18.
Now that I look over your work again, I agree with what you've done except I used 0.708 instead of 0.71. I think you just made a math error because
(x/0.153-x) = 0.708 is right to that point. It's the 128 that isn't right.
I keep getting the last part of this question wrong.
Strong base is dissolved in 765 mL of 0.200 M weak acid (Ka = 4.63 x 10^-5) to make a buffer with a pH of 4.18. Assume that the volume remains constant when the base is added. HA(aq) + OH-(aq) -. H2O(l) + A-(aq). I found that the pKa of the acid = 4.33 and the number of moles of acid initially present is 0.153 mol HA. [A-]/[HA] = 0.71 which are correct but i can't find how many moles of strong base were initially added!
(x/0.153-x) = .71 ----> 0.128 mol of OH is incorrect
and 0.153 [which i really thought would be correct] is incorrect as well!
y = 0.71(0.153 mol - y)
y = 0.153 mol
PLEASE help me!! this problem is frustrating me
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