Question

1) If the Ka of a monoprotic weak acid is 2.1

Answer 1

a)

Let HA be the monoprotic acid and let A- be its conjugate base.

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

Which has an equilibrium expression of: Ka = ([H+][A-]) / [HA], where Ka is the equilibrium constant and the value in brackets is concentration.

Make an ICE table. Initial concentration of HA is 0.45. Initial concentration of A- and H+ is zero. As time proceeds and the reaction shifts towards equilibrium, HA will lose x M and H+ and A- will gain x.

@ Equilibrium
[HA] = 0.48 - x
[A-] = x
[H+] = x

Ka = ([H+][A-]) / [HA]
2.1 * 10^-6 = ((x)(x)) / (.48 - x)
2.1(.48 - x) = x^2

Using algebra (not recommended), a calculator, or a solver function, we find that x = 9.71 * 10^-4 (ignore the negative value, negative concentrations don't work). We defined x to be the concentration of H+ and A- at equilibrium. We only care about H+ for pH.

pH = -log(9.71 * 10^-4)
pH = 3.01

b)

[OH-] = (Kb * Co)^1/2 only for weak base. as for weak acid is [H+]=(Ka * Co)^1/2

SO CONCENTRATION OF OH- IS 5.648* 10^-4 M
pOH = -log(5.648*10^-4)
pOH=3.248 = 3.25 (approx)
pH + pOH = 14
so pH= 10.75