Chemistry Buffer/titration problem

Question

Hello,

I am having trouble with this problem. I think I have part a) using the Henderson-Hasselbalch equation but part b) and c) are confusing me. I know I need to set the pH=7 and rearrange the HH equation but after that I do not know what to do.

I've changed the numbers slightly from the actual problem in my homework but here is the problem:

370.0 mL of buffer is known to contain 0.325 M CH3COOH and 0.630 M NaCH3COO. a. Determine the initial pH of this buffer b. Determine the volume (mL) of 0.250 M HCl required to neutralize the buffer. c. Determine the volume (mL) of 0.250 M NaOH required to neutralize the buffer. ka=1.76E-5

Thank you

Answer 1

For parts (b) and (c) "to neutralize the buffer" doesn't necessarily mean to get to pH 7, but probably means to get the solution to the point where there is no more buffering capacity. Otherwise, you couldn't use both HCl AND NaOH to "neutralize" the buffer and end up with pH = 7, right?

(b) For HCl: (it reacts with the CH3COO- to produce CH3COOH)

moles CH3COO- = 0.370 L x 0.630 mol/L = 0.233 moles

volume (ml) HCl needed = (x L)(0.250 mol/L)(1000 ml/L) = 0.233 moles

x = 932 mls HCl (this converts ALL CH3COO- to CH3COOH and no more buffering capacity)

(c) For NaOH: (it reacts with the CH3COOH to produce CH3COO-)

moles CH3COOH = 0.370 L x 0.325 mol/L = 0.120 mol

volume (ml) NaOH needed = (x L)(0.250 mol/L)(1000 ml/L) = 0.120 mol

x = 481 mls NaOH (this converts ALL CH3COOH to CH3COO- and no more buffering capacity)

If the question is asking you to adjust the pH to 7, then solve the HH equation for the ratio [CH3COO-]/[CH3COOH] as follows:

7.00 = 4.75 + log [CH3COO-]/[CH3COOH]

log [CH3COO-]/[CH3COOH] = 2.25

[CH3COO-]/[CH3COOH] = 178

This can only be accomplished by the addition of NaOH. Addition of HCl will lower the pH.

(0.233 + x)/(0.12-x) = 178

179x = 21.2

x = 0.118 = moles CH3COOH to be converted to 0.118 moles CH3COO-

moles NaOH = 0.118

volume NaOH = (x L)(0.25 mol/L)(100 ml/L) = 0.118 moles

x = 472 mls NaOH