ph=______
For a weak acid, HA, the equilibrium is:
HA <--> H+ + A-
Ka = [H+][A-]/[HA] = 6.1X10^-6
Let x = [H+] = [A-], and [HA] = 0.28 - x. Because Ka is small, x will probably be much smaller than 0.28 and so can be ignored. Then,
Ka = 6.1X10^-6 = x^2 / 0.28
x = [H+] = 1.3X10^-3 (x is indeed small compared to 0.28, so ignoring it was fine)
pH = - log (1.3X10^-3) = 2.88
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Verified answer
For a weak acid, HA, the equilibrium is:
HA <--> H+ + A-
Ka = [H+][A-]/[HA] = 6.1X10^-6
Let x = [H+] = [A-], and [HA] = 0.28 - x. Because Ka is small, x will probably be much smaller than 0.28 and so can be ignored. Then,
Ka = 6.1X10^-6 = x^2 / 0.28
x = [H+] = 1.3X10^-3 (x is indeed small compared to 0.28, so ignoring it was fine)
pH = - log (1.3X10^-3) = 2.88