Posted by **Kevin ** on Thursday, October 20, 2011 at 1:21pm.

so i had to calculate the pH of a 100ml solution of .10M acetic acid (ka=1.8x10(-5) I came out to 2.87. But then, they want the pH if 50ml HCL is added. So i thought...Ka = x2/.15M, therefore the square root of .15M x 1.8x10(-5) = .00164(concentration). So -log(.00164) = 2.78. Can you verify this, or tell me where i went wrong?

- Chemistry -
**DrBob222**, Thursday, October 20, 2011 at 3:52pm
2.87 for pH of 0.1M HAc is right.

For HCl + HAc first, what is the concn of the HCl added? If we assume it is 0.1M (and it may not be that at all but you can adjust).

(HAc) = 0.1M x (100/150) = 0.0667

(H^+) = x from HAc and 0.0333 from HCl (that is 50 mL x 0.1M diluted to 150 mL or 0.1 x 50/150 = 0.0333M).

(Ac^-) = x

Ka = (H^+)(Ac^-)/(HAc)

1.8E-5 = (x+0.0333)(x)/(0.0667-x)

I solved this and x is negligible compared to 0.0333 or 0.0667 so pH is determined by the HCl so (H^+) = 0.0333 and get pH from that or about 1.5 or so.

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