chemistry

A buffer is prepared using acetic acid, CH3COOH, (a weak acid, pKa = 4.75) and sodium acetate, CH3COONa (which provides acetate ions, the conjugate base), according to the following proportions:

Volume of CH3COOH(aq): 100.0 mL
Concentration of CH3COOH(aq): 1.151 M

Volume of CH3COONa(aq): 113.0 mL
Concentration of CH3COONa(aq): 1.270 M

Use the Henderson-Hasselbalch equation (below) to calculate the pH of the buffer solution.
pH=pKa+log[A−][HA]

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1. equation is pH=pKa+log[A−]/[HA]?

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2. yes

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3. Plug in the numbers and calculate it.

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DrBob222
4. is it just = 4.75+log(1.270/1.151)?

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5. Yes, it's just that.

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DrBob222
6. you dont need to take the volume into consideration?

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7. The formula is pH = pKa + log (base)/(acid)/ Concn is in mols/L which is M
so you substitute the MOLAR CONCENTRATIONS and that's it.

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DrBob222
8. i see, thanks!

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9. i think its this:
Volume of final solution = (100.0 + 113.0) mL = 213.0 mL

[CH₃COOH] = (1.151 M) × (100.0/213.0) = 0.5404 M
[CH₃COO⁻] = (1.270 M) × (113.0/213.0) = 0.6738 M

Henderson-Hasselbalch equation:
pH = pKa + log([CH₃COO⁻]/[CH₃COOH])
pH = 4.75 + log(0.6738/5404)
pH = 4.85

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10. yes, youre right vace

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11. I hate to say it but you guys are right and I'm wrong. The volume DOES make a difference although not in a big way. My mistake is this.
I looked at the divisor of 213 and thought dividing by a common number left concentrations the same so the concentrations would be the same. However, that isn't true since the initial volumes were different. Sorry about that. I just didn't think it through.

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DrBob222
12. no worries, your help still means a lot!

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