CHEMISTRY_ Using the expression Ka=[H+][A−] / [HA], explain how to determine which solution has the

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Question Answered step-by-step (a) Using the expression K a =[H+][A−] / [HA], explain how to...

Image transcription text Acid Ka HF 6.7 x 10 -4 HC,H3 02 1.8 x 10 -5 (a) Using the expression Ka=[H+][A−] / [HA], explain how to determine which solution has the lower pH , 0.10 M HF(aq) or 0.10 M HC2H3O2(aq). Do not perform any numerical calculations. (b) Which solution has a higher percent ionization of the acid, a 0.10 M solution of HC2H3O2(aq) or a 0.010 M solution of HC2H3O2(aq)? Justify your answer including the calculation of percent ionization for each solution.

(c) A student titrated 50.0mL of a 0.10 M solution of a certain weak acid with NaOH(aq). The results are given in the graph above. (i) What is the approximate pKa of the acid?

(ii) What specific value from the graph is needed, in addition to the information in part (c) above, to calculate the molar concentration of the NaOH(aq)? (iii) Identify a pH value between 2.5 and 7.5 at which the concentration of the weak acid being titrated is less than the concentration of its conjugate base. (d) In another experiment, the student titrated 50.0mL of 0.100 M HC2H3O2 with 0.100 M NaOH(aq). Calculate the pH of the solution at the equivalence point.

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Step-by-step explanation (a) Ka=[H+][A−]/[HA] which solution has the lower pH, 0.10MHF(aq) or 0.10MHC2H3O2(aq) : Higher value of Ka indicates greater ionization thus more protons in solution, higher H+ concentration in solution means lower pH value.

Since HF have higher Ka have, thus HF solution will have lower pH. (b) Which solution has a higher percent ionization of the acid, a 0.10M solution of HC2H3O2(aq) or a 0.010M solution of HC2H3O2(aq) : Ka for this acid : 1.8*10^-5 Ka=[H+][A−]/[HA] = 1.8*10^-5 0.01 M solution will have higher % of ionization, than 0.1 M solution, Ka = [H+][A−]/(0.01 -x)= 1.8*10^-5 [H+] = [A-] = x ===> x2 ~1.8*10^-7 thus, for 0.01 M : x = 0.000424 M % ionization: 4.24 % And, [H+][A−]/0.1-x = 1.8*10^-5 [H+] = [A-] = x ===> x2 ~ 1.8*10^-6 For 0.1 M , x = 0.00134 % Ionization : 1.34 % C. student titrated 50.0mL of a 0.10M solution of a certain weak acid with NaOH(aq). The results are given in the graph above. (i) What is the approximate pKa of the acid : Initial pH ~2.1 ==> [H+] = 10^-2.1 = ( Ka * 0.10 )^1/2 Thus, Ka ~ 6.3*10^-4 ==> pKa = 3.2 (ii) What specific value from the graph is needed, in addition to the information in part (c) above, to calculate the molar concentration of the NaOH(aq) : volume of NaOH needed for equivalence point, or equivalence point pH.

(iii) Identify a pH value between 2.5 and 7.5 at which the concentration of the weak acid being titrated is less than the concentration of its conjugate base : During titration before equivalence point solution behave as buffer system of weak acid and its conjugated base, when pH ~ pKa , concentration of both weak acid and its conjugated base is equal thus , as pH crosses pKa value concentration of weak acid becomes lesser than concentration of its conjugated base. Thus, answer is pH 3.5 and above. (d) In another experiment, the student titrated 50.0mL of 0.100MHC2H3O2 with 0.100MNaOH(aq). Calculate the pH of the solution at the equivalence point.

At equivalence point pH is determined by conc . Of conjugated base as it generates OH- ions : CH3COO- + H2O <==> CH3COOH + OHFinal V = 100 ml [CH3COO-] = 50 ml *0.100 M / 100 ml = 0.05 M kb = 10^-14/Ka = 5.56*10^-10 [OH-] = (Kb* 0.05)^1/2 = 1.67*10^-6 M pH =14 - pOH = 8.23