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
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])
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.
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.
To calculate the pH before the addition of HCl (part A), you need to use the Henderson-Hasselbalch equation:
pH = pKa + log{[C2H3O2^-]/[HC2H3O2]}
In this equation, pKa is the dissociation constant of acetic acid, [C2H3O2^-] is the concentration of sodium acetate, and [HC2H3O2] is the concentration of acetic acid.
From your data, the concentration of sodium acetate is given as 0.1 M. However, the concentration of acetic acid is not directly given. You mentioned that the volume of acetic acid used is 20 ml, but that is not sufficient information to determine the concentration.
To calculate the pH after the addition of HCl (part B), the reaction between HCl and NaC2H3O2 converts half of the C2H3O2- ions to HC2H3O2. This means the concentration of C2H3O2- is reduced by half, while the concentration of HC2H3O2 is increased by half. Thus, you can use the same Henderson-Hasselbalch equation, but with modified concentrations:
pH = pKa + log{[C2H3O2^-]/[HC2H3O2]}
However, in this case, [C2H3O2^-] is halved since half of the C2H3O2- ions are converted to HC2H3O2. So the equation becomes:
pH = pKa + log{(0.5*[C2H3O2^-])/([HC2H3O2])}
To complete the calculations and find the pH values, you need to look up the pKa value of HC2H3O2. The pKa value represents the acid dissociation constant and can be found in reference books or online databases.
Once you know the pKa value, you can substitute it into the Henderson-Hasselbalch equation to calculate the pH before and after the addition of HCl.