posted by Saira on .
A beaker with 150 mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 5.60 mL of a 0.280 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.
[CH3COOH] + [CH3COO-]= 0.100 M
pH = pKa + log [CH3COO-] / [CH3COOH]
5.00 = 4.760 + log [CH3COO-] / [CH3COOH]
10^0.24 =1.74 = [CH3COO-]/ [CH3COOH]
I have an idea how to solve it but i dunt know how to get the M of CH3COO AND CH3COOH
So far so good. To save time in typing, I'll call CH3COO- (acetate) = B (for base) and CH3COOH(acetic acid) = A (for acid).
So what you have is B/A = 1.74 and we can rearrange that to
B = 1.74 x A (I would carry that out to one more place; i.e., instead of 1.74 I would go with 1.738.
Now you know that
B + A = 0.1 so substitute for B from above to obtain
1.738A + A = 0.1
and solve for A, then B.
That's how you obtain A and B (concn in moles/L = M). Then in short, the rest of the problem is
How much acid is in the buffer of 5.00? That is 150 mL x (A) = ??
How much base is in the buffer of 5.00?
That is 150 mL x (B) = ??
Now, when you add the 0.280 M HCl what is happening?
Base + HCl ==> acid + Cl^-
So how much acid are you adding? That is 0.280 M x 5.60 mL = ??= I'll say x.
So you add x to acid and subtract x from the base, then reset pH = pKa + log (B/A) and solve for the new pH.
I did a quickie calculation and obtained 4.81 or so. You need to redo the whole thing. Check my work. Does that answer make sense? Yes, because it was pH = 5.00 and we added HCl, a strong acid to it, so it should become more acid and 4.81 is more acid than 5.00.