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What is pH when 0.5 m HAC (acetic acid k_a = 1.8 times 10^-5) is allowed to go to equilibrium? HAC + H_2O = AC^- + H_3O^+
Expert Answer
Answer
HAC dissociates as:
HAC + H2O —–> H3O+ + Ac-
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Question & Answer: What is pH when 0.5 m HAC (acetic acid k_a = 1.8 times 10^-5) is allowed to go to…..
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0.5 0 0
0.5-x x. x
Ka = [H+][Ac-]/[HAC]
Ka = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((1.8*10^-5)*0.5) = 3*10^-3
since c is much greater than x, our assumption is correct
so, x = 3*10^-3 M
so.[H+] = x = 3*10^-3 M
use:
pH = -log [H+]
= -log (3*10^-3)
= 2.52
Answer: 2.52
