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CHM-1: × ' ..,' owLv2 1 Online teaching and x enow.com,/ilrn/takeAssignment/takeCovalentActivity,do?locator-assignment-take&takeAssignmentSessic; Academica 2017 INTERACTIVE EXAMPLE Determining Ka from the Solution pH d, HA, is a monoprotic acid. A solution that is 0.270 M in HA has a pH of2.400 at 25°C. HA(aq) + H2O(l) ,0'(aq) + A-(aq) What is the acid-ionization constant, Ka, for this acid? What is the degree of ionization of the acid in this solution? Degree of ionization- Submit Hide Tutor Steps TUTOR The next step is to construct a table showing the initial concentrations, the change in concentrations, and t sometimes called an ICE table. We call the change in concentration of H3O+ and then use that value to all species. What will be the equilibrium concentration of A, in terms of x? PROBLEM MAP pH 2.400

Explanation / Answer

a)

we have below equation to be used:

pH = -log [H+]

2.4 = -log [H+]

log [H+] = -2.4

[H+] = 10^(-2.4)

[H+] = 3.981*10^-3 M

Lets write the dissociation equation of HA

HA -----> H+ + A-

0.27 0 0

0.27-x x x

[H+] = x = 3.981*10^-3 M

Ka = [H+][A-]/[HA]

Ka = x*x/(c-x)

Ka = 3.981*10^-3*3.981*10^-3/(0.27-3.981*10^-3)

Ka = 5.96*10^-5

Answer: Ka = 5.96*10^-5

b)

degree of ionisation = x/c

= (3.981*10^-3)/0.27

= 0.0147

Answer: 0.0147

Feel free to comment below if you have any doubts or if this answer do not work

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