Part 2: Use the theoretical value (Ka of acetic acid alone 1.76x10-5) to calcula

Question

Part 2: Use the theoretical value (Ka of acetic acid alone 1.76x10-5) to calculate the pH of acetic acid alone with an l.?.?. table and of the two buffer solutions using the Henderson- Hasselbalch equation. Compare your measured and calculated pH's in Table 2. Remember, the % error equation is found in Appendix 3. Using I.C.E tables and measured pH, calculate the Ka for acetic acid alone, and acetic acid with both10.0 mL and 1.0 mL portions of sodium acetate and write the Ka's in Table 6. Note: do NOT assume x is negligible. In this case the x is given from the measured pH because x [H3O]. Table 6: Acid-Dissociation Constants for Acetic Acid Buffer solutions Ka of 1.0 M acetic acid alone Ka of buffer: 10 mL acetic acid 10 mL sodium acetate Ka of buffer: 10 mL acetic acid1 mL sodium acetate

Explanation / Answer

Ka of acetic acid = 1.76*10-5, pKa= -log Ka= 4.74

Acetic acid undegoes iomization as CH3COOH+ H2O ----------->CH3COO- + H3O+

Ka= [CH3COO-][H3O+]/[CH3COOH]

preparing the ICE table

component           initial concentration (M)                         Change in concentration(M)           Eq.concentration(M)

CH3COOH                1                                                              -x                                                     1-x

CH3COO-                  0                                                               x                                                       x

H3O+                         0                                                              x                                                         x

hence Ka= x2/(1-x)= 1.76*10-5, when solved using excel, x=[H3O+]= 0.004186, pH= -log (0.004186)= 2.378

measured pH= 2.51, % error= 100*(2.51-2.378)/2.51=5.25%

2. moles of CH3COONa = Molarity* volume in liters= 1*10/1000 =0.01 ,moles of acetic acid= 0.01

volume of mixing =10ml+10ml=20ml= 20/1000L. =0.02L

concentrations after mixing :[CH3COONa]= [CH3COOH] = 0.01/0.02 =0.5

From Henderson-Hasselbalch equation, pH= pKa+ log{[A-]/[HA]}=4.74+log (0.5/0.5)= 4.74

measured value =4.92, % error= 100*{(4.92-4.74)/4.92}=3.66%

moles : CH3COONa= 1*1/1000 =0.001 and CH3COOH= 1*10/1000=0.01

volume after mixing= 10+1=11ml, =11/1000L= 0.011L

concentrations : [CH3COONa]= 0.001/0.011=0.091, [CH3COOH] =0.01/0.011 =0.91

now from Henderson-Hassel balch equation

pH= 4.74+log ( 0.091/0.91)= 3.74

% error= 100*(3.77-3.74)/3.77 =0.795%

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