3)
use:
pH = -log [H+]
1.79 = -log [H+]
[H+] = 1.622*10^-2 M
HA dissociates as:
HA -----> H+ + A-
0.25 0 0
0.25-x x x
Ka = [H+][A-]/[HA]
Ka = x*x/(c-x)
Ka = 1.622*10^-2*1.622*10^-2/(0.25-1.622*10^-2)
Ka = 1.125*10^-3
Answer: 1.12*10^-3
Only 1 question at a time please
A solution contains 0.250 MHA(Ka-1.0 x 106and 0.45 M NaA. What is the pH after 0.30 mole of HCl is added to 1.00 L of this solution? a. 7.44 .2.10 98.56 8.5.44 e.0.52
What concentration of acetic acid (Ka= 1.80x10^-5) has the same pH as that of 5.61x10^-3 M HCl?
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