# Chemistry

posted by on .

I actually posted this up before and Gk helped out but i don't understand the steps what do i enter for the pka for the first one?

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I did an experiment on Buffers:

In a polystyrene beaker, mix 20 ml of 0.1M Acetic acid ad 25 ml of 0.1 M sodium acetate and immediately measure the pH. Remove the electrode and add 5 ml of 0.1 M HCL to this buffer. Stir the solution and measure the pH.

This is my data:

Conc.of acetic acid 0.1 M- volume 20mL
Conc.of sodium acetate 0.1M-volume 25ml
Conc. of Hydrochloric acid 0.1M-volume 5 ml

Buffer Solution- pH measured- 4.65
pH calculated ?

Buffer solution + HCL -pH measured- 4.43
pH calculated-?

A. So basically i need to show calculation for the calculated pH before the addition of HCl

B. Show the calculations for the calculated pH after the addition of HCl
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Responses

* Chemistry - GK, Thursday, July 16, 2009 at 5:16pm

The equilibrium in both mixtures is:
HC2H3O2(aq) <=> H+(aq) + C2H3O2^-(aq)

For the first mixture use the Henderson-Hasselbalch Equation to get the pH:
pH = pKa + log{[C2H3O2^-]/[HC2H3O2]}
NOTE: [C2H3O2^-] = molarity of NaC2H3O2
pH = pKa + log{1}

In the second mixture, reaction between HCl and NaC2H3O2 converts half of the C2H3O2- ions to acid, HC2H3O2, so,
pH = pKa + log{1/2}

Look up the pKa of HC2H3O2 and complete the calculations

• Chemistry - ,

This is what i have so far...The Ka is 1.8*10^-5

so pKa is= -log(1.8*10^-5)
= 4.74

pH= pKa+log ([C2H3O2^-]/[HC2H3O2])
= 4.47+ log ([C2H3O2^-]/[HC2H3O2])

• Chemistry - ,

pKa is the pKa for acetic acid. Since Ka for acetic acid is about 1.8 x 10^-5, the pKa is -log Ka = about 4.75 or so. You need to look up Ka for acetic acid in your text or notes and take the negative log of that value.

• Chemistry - ,

I think you had two calculations to make; one of the original pH of the solution and the second one after the adition of HCl to the mixture. Look at the problem. It appears to me that the initial concentration of HC2H3O2 is 0.1 M and the initial concn of C2H3O2^- is 0.1 so that is the log (0.1/0/0.1) = log 1 and that is zero. GK did that part for you.