I need d answered please. I answered a-c already.

Post by **answerhappygod** » Mon Jul 11, 2022 2:03 pm

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5. (54 points) Calculate the pH at the following points in the titration of 20.00 ml of 0.100 M CH3COOH with 0.100 M KOH. The Ka of CH3COOH is 1.8x10-5 a. Before the addition of any KOH CH3COOH -----> H+ + CH3COO- 0.100 0.1-x Ka = [H+][CH3COO-]/[CH3COOH] Ka = x*x/(c-x) Ka = x*x/(c) so, x = (Ka*c) x = ((1.8*10^-5)*0.1) = 1.342*10^-3 Ka = x*x/(c-x) 1.8*10^-5 = x^2/(0.1-x)| 1.8*10^-6 1.8*10^-5 *x=x^2 x^2 + 1.8*10^-5 *x-1.8*10^-6=0 (ax^2+bx+c=0) a = 1 b = 1.8*10^-5 c = -1.8*10^-6 x = {-b + (b^2-4*a*c)}/2a x = {-b-(b^2-4*a*c)}/2a b^2-4*a*c = 7.2*10^-6 x = 1.333*10^-3 and x = -1.351*10^-3 x = 1.333*10^-3 pH = -log [H+] = -log (1.333*10^-3) = 2.8753 = 2.88
b. After the addition of 5.00 ml of 0.100 M KOH M(CH3COOH) = 0.1 M V(CH3COOH) = 20 mL M(KOH) = 0.1 M V(KOH) = 5 mL mol(CH3COOH) = M(CH3COOH)* V(CH3COOH) mol(CH3COOH) = 0.1 M * 20 mL = 2 mmol mol(KOH) = M(KOH) * V(KOH) mol(KOH) = 0.1 M * 5 mL = 0.5 mmol mol(CH3COOH) = 2 mmol mol(KOH) = 0.5 mmol 0.5 mmol of both will react excess CH3COOH remaining = 1.5 mmol Volume of Solution = 20 + 5 = 25 mL [CH3COOH] = 1.5 mmol/25 mL = 0.06M [CH3COO-] = 0.5/25 = 0.02M Ka 1.8*10^-5 pKa = -log (Ka) = -log(1.8*10^-5) = 4.745 pH =pKa + log {[conjugate base]/[acid]} = 4.745+ log {2*10^-2/6*10^-2} = 4.268 = 4.27
C. After the addition of 10.00 ml of 0.100 M KOH M(CH3COOH) = 0.1 M V(CH3COOH) = 20 mL M(KOH) = 0.1 M V(KOH) = 10 mL mol(CH3COOH) = M(CH3COOH) * V(CH3COOH) mol(CH3COOH) = 0.1 M * 20 mL = 2 mmol mol(KOH) = M(KOH) * V(KOH) mol(KOH) = 0.1 M * 10 mL = 1 mmol mol(CH3COOH) = 2 mmol mol(KOH) = 1 mmol excess CH3COOH remaining = 1 mmol volume of Solution = 20 + 10 = 30 mL [CH3COOH] = 1 mmol/30 mL = 0.0333M [CH3COO-] = 1/30 = 0.0333M acid is CH3COOH conjugate base is CH3COO- Ka 1.8*10^-5 pKa = -log (Ka) = -log(1.8*10^-5) = 4.745 pH = pKa + log {[conjugate base]/[acid]} = 4.745+ log {3.333*10^-2/3.333*10^-2} = 4.745 =4.75
d. After the addition of 20.00 ml of 0.100 M KOH