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
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
2 answers
only to correct that conce of acetate is not equal to that of acetic acid molatrity of HC2H3O2=2 molarity of NaC2H3O2=2.5 so pH=pKa+log2/2.5 . =pKa+log0.8
3 answers