Question

Part - USING PA to Determine the value of K. for Aceto Adid, CH.COOH ) .Write the chemical equation describing the equilibrium reaction between acetic acid and water CH, cool caq) + H₂O) H₂O trac) + CHycoo Taq) 1.8x10-5 3.73 . Complete the following table. Use the known value of K for acetic acid from your textbook to determine the percentage error in your measured Kvalue for each solution 1.0-M CH.COOH 0.010-M CH.COOH Measured pH 2.94 1.148*10M 1.86 x 10-4M [CH,COOL 1.148X10-3M 1 1,86810 "M [CHCOOH) 9.98 x10. I'M 9 9.81 x 10 BM K (calculated) 3.52 x 10 K (textbook) 1.8x10-5 1.8 X1D-5 Percent error 92.67% 80.44 % • Show your calculations (using an equilibrium or ICE table for obtaining the value of K, for the solution (only): CH₃COOH cag) + H₂O H₂O + cag) + CH₂(coo-Xag) 1 .010 M +x C -x x = 1.86810-4 M X = 1,86x10 - DOM-X =9.84X10-3M [Hot] = 10 -3.73 [4,0*) - 1.962x10-4 Kaz (1.86610-4 M) (1.86x10-44) 9.81810-3M Ka - 3.525 x 10-6 % = $3.525 X10 % - 1.8x10-5. xloo 1.8x10-S = 80,44%

It appears that my percent error is very large. Am I calculating this correctly?

Thank you

Answer 1

a) Ka = [H₃O]⁺ [CH₃COO^-]/[CH₃COOH]

[H₃O]⁺ = [CH₃COO^-] = 10^-pH = 10^-2.94 = 0.001148 M

[[CH₃COOH] = initial concen - [CH₃COO^-] = 1-0.001148 = 0.998852

Ka = 0.001148²/0.998852 = 0.00000132/0.998852 = 1.32 x 10^-6

Literature value = 1.8 x 10^-5

Percent error = [actual-literature]/literature ) x 100 = ([0.00000132-0.000018])/0.000018) x 100 = 0.00001668/0.000018) x 100 = 92.7%

b) Ka = [H₃O]⁺ [CH₃COO^-]/[CH₃COOH]

[H₃O]⁺ = [CH₃COO^-] = 10^-pH = 10^-3.73 = 0.000186 M

[[CH₃COOH] = initial concen - [CH₃COO^-] = 0.01-0.001148 = 0.009817

Ka = 0.000186²/0.009817 = 0.0000000346/0.009817 = 3.52 x 10^-6

Literature value = 1.8 x 10^-5

Percent error = [actual-literature]/literature ) x 100 = ([0.00000352-0.000018])/0.000018) x 100 = 0.00001448/0.000018) x 100 = 80.45%

Yes your calculation is correct probably measured pH were not correct