Focus on the buffer system of acetic acid and sodium acetate:
HC2H3O2 / C2H3O2⁻
acetic acid + acetate ion
First, identify which component is the conjugate acid, and which is the conjugate base. Write the balanced equation for the buffer – it is easiest to write the acid on the left side of the equation. Round all final calculated Ka values to three significant figures, and your calculated pKa to two decimal places.
Recall: pKa = –log(Ka).
Three different solutions were prepared using 0.10 M stock solutions of NaC2H3O2 and HC2H3O2.

1. pH of a Buffer, and Calculation of Ka.
A solution was prepared by measuring 15.0 mL of the acid component and 15.0 mL of the conjugate base of your buffer into a 100-ml beaker. The pH of the resulting solution was 4.81. Calculate Ka, and then pKa for the acid from this pH.

4.81= -log 0.05/0.05 Ka = 10^(-pKa) = 10^4.81
pKa = 4.81 Ka = 1.55x10^(-5)

2. pH of a Diluted Buffer.
Distilled water was added to Solution 1 above until the volume was double its original volume (here. V_f = 60 mL). The pH of the resulting solution was 4.83, which did not change significantly from Solution 1. Does this make sense? Address this point in your discussion /conclusion

3. pH of a Different [HA]/[A⁻] Ratio Buffer, Calculation of Ka.
A solution was prepared by measuring 2.0 mL of the stock acid component and 20.0 mL of the stock solution containing its conjugate base. The pH of the resulting solution was 5.73. Calculate K. and then DK for the acid from this pH.

C_acid = (0.10Mx 2 ml)/22 ml = 9.09x10^(-3)M
pKa = 5.73 -log ([19.09x10^2]/[9.09x10^-3] = 5.73 -1 = 4.73
C_base=(0.10M x 20 ml)/22 ml = 9.09x10^2 M
Ka = 10^(-4.73) = 1.86x1045 11

Chemical Equation of Acid/Base Equation for buffer system reacting with H2O:
HC2H3O2 +H2O⁻ → H3O⁺ +C2H3O2⁻

Calculation:
Ensure proper units and significant figures for all reported values.
Whenever possible, type out the equations that you are solving with the variables first, and then with values inserted, then report final numerical answer.
1. Provide the calculation of the reaction quotient, Q. for Solution 3.
Q = [lead ion] x [iodide ion]²
Q= 0.001612 * 0.0016 = 4.096E-9

2. Use an ICE Table or the Henderson-Hasselbalch equation Provide the calculation of the K. and then pKa, for the acidic Solution 1.
pH = pka+log 10[conjugate base]/[weak acid]
[acid]-0.05M and [salt]-0.05M pH=pKa + log [NaC2H3O2[HC3H3O2]
4.81= -log 0.05/0.05 Ka=10^(-pKa) =10^(4.81)
pKa = 4.81 Ka= 1.55x10^5

3. Use an ICE Table or the Henderson-Hasselbalch equation.
Provide the calculation of the Ka, and then pKa, for the acidic Solution 3.
4. Use an ICE Table or the Henderson-Hasselbalch equation Provide the calculation of the volumes necessary to prepare for the buffered Solution 3.