Question
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
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
Answers
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
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
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
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