If the Ka of a monoprotic weak acid is 3.6 × 10^-6, what is the pH of a 0.35 M solution of this acid?
A certain weak base has a Kb of 7.30 × 10-7. What concentration of this base will produce a pH of 10.34?_____M
A certain weak base has a Kb of 7.30 × 10-7. What concentration of this base will produce a pH of 10.34?_____M
-
HA ===> H+ + A-
Ka = [H+][A-]/[HA] = 3.6 x10^-6
Let [H+] = x. Then [A-] = x and [HA] = 0.35 - x. But x is so small that we can neglect it and say that[HA] = 0.35.
Ka = (x)(x)/(0.35) = 3.6 x 10^-6
x^2 = 1.26 x 10^-6, so x = 1.12 x 10^-3
pH = -Log[H+], and Log(1.12x10^-3) = -2.95, so pH = 2.95
BOH ===> B+ + OH-
Kb = [B+][OH-]/[BOH] = 7.30 x 10^-7, so [BOH] = (7.30x10^-7)/[B+][OH-]
pH + pOH = 14.00, and pH = 10.34, so pOH = 14.00 - 10.34 = 3.66
pOH = 3.66, so Log[OH-] = -3.66, and [OH-] = 2.19 x 10^-4
If [OH-] = 2.19 x 10^-4, then [B+] = 2.19 x 10^-4 and
[BOH] = (7.30x10^-7)/(2.19x10^-4)(2.19x10^-4) = 15.2 M
Ka = [H+][A-]/[HA] = 3.6 x10^-6
Let [H+] = x. Then [A-] = x and [HA] = 0.35 - x. But x is so small that we can neglect it and say that[HA] = 0.35.
Ka = (x)(x)/(0.35) = 3.6 x 10^-6
x^2 = 1.26 x 10^-6, so x = 1.12 x 10^-3
pH = -Log[H+], and Log(1.12x10^-3) = -2.95, so pH = 2.95
BOH ===> B+ + OH-
Kb = [B+][OH-]/[BOH] = 7.30 x 10^-7, so [BOH] = (7.30x10^-7)/[B+][OH-]
pH + pOH = 14.00, and pH = 10.34, so pOH = 14.00 - 10.34 = 3.66
pOH = 3.66, so Log[OH-] = -3.66, and [OH-] = 2.19 x 10^-4
If [OH-] = 2.19 x 10^-4, then [B+] = 2.19 x 10^-4 and
[BOH] = (7.30x10^-7)/(2.19x10^-4)(2.19x10^-4) = 15.2 M