ML>katifund.org: offered the concentration and also the Ka, calculation the percent dissociationGiven the concentration and the Ka, calculation the percent dissociationReturn to the acid Base menuReturn to a listing the many types of acid base problems and their solutionsProblem #1: calculation the percent dissociation of a weak mountain in a 0.050 M HA solution. (Ka = 1.60 x 10-5)Solution:1) calculate the :1.60 x 10-5 = <(x) (x)> / 0.050x = 8.9443 x 10-4 M2) division the by the concentration, then multiply by 100:(8.9443 x 10-4 M / 0.050 M) x 100 = 1.79% dissociatedProblem #2: A specific weak acid, HA , has a Ka worth of 9.2 x 10¯71) calculation the percent dissociation of HA in a 0.10 M solution.2) calculation the percent dissociation of HA in a 0.010 M solution.3) calculate the percent dissociation that HA in a 0.0010 M solution.Solution to component one:Step #1: calculate the :9.2 x 10¯7 = <(x) (x)> / (0.10 - x)neglect the minus xx = 3.03315 x 10¯4 M(note the I kept some guard digits, I"ll round off the final answer.)Step #2: division the by the concentration, then multiply by 100:(3.03315 x 10¯4 M / 0.10 M) x 100 = 0.303% dissociatedSolution to component two:Step #1: calculate the :9.2 x 10¯7 = <(x) (x)> / (0.010 - x)neglect the minus xx = 9.591663 x 10¯5 M(note that I preserved some safety digits, I"ll round turn off the final answer.)Step #2: division the by the concentration, climate multiply through 100:(9.591663 x 10¯5 M / 0.010 M) x 100 = 0.959% dissociatedSolution to component three:Step #1: calculate the :9.2 x 10¯7 = <(x) (x)> / (0.0010 - x)neglect the minus xx = 3.03315 x 10¯4 M(note that I preserved some guard digits, I"ll round off the final answer.)Step #2: divide the through the concentration, climate multiply by 100:(3.03315 x 10¯5 M / 0.0010 M) x 100 = 3.03% dissociatedYes, as the solution becomes an ext dilute, the percent dissociation increases. However, as the equipment becomes more and much more dilute, the above method breaks down.


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(At 0.00000010 M, the percent dissociation is 303%.) This is due to the fact that the is becoming larger and larger compared to the concentration of the acid. This means that the neglecting "x" approximation introduces much more and an ext error.This means that, as the dilution increases, us must transition to more sophisticated calculational approaches which are beyond the scope of this internet site.Problem #3:
calculation the degree of ionization the acetic acid in the following solutions:solution 1 : 0.10 M HC2H3O2solution 2 : 5 mL 0.10 M HC2H3O2 + 5 mL H2Osolution 3 : 1 mL 0.10 M HC2H3O2 + 99 mL H2OSolution to part one:1) calculation the : = √(Ka time concentration) = √(1.77 x 10¯5 time 0.10) = 0.00133 M2) divide the by the concentration, then multiply through 100:(0.00133 /0.1) x 100 = 1.33%Solution to part two:1) calculation the : = √(Ka time concentration) = √(1.77 x 10¯5 time 0.050) = 0.000941 M2) division the by the concentration, then multiply by 100:(0.000941 /0.05) x 100 = 1.88%Comment: note exactly how the systems was diluted native 5 mL come 10 mL, for this reason cutting the concentration in half.Solution to component three:1) calculation the brand-new molarity utilizing M1V1 = M2V2:(0.10) (1) = (x) (100)x = 0.0010 M2) calculate the : = √(Ka times concentration) = √(1.77 x 10¯5 time 0.0010) = 0.000133 M3) division the by the concentration, then multiply through 100:(0.000133 /0.0010) x 100 = 13.3%Return to the mountain Base menuReturn come a listing the many species of acid base problems and also their solutions