Weak base in water, (Salt in Water) ?

Question

What is the pH of a lithium cyanide solution prepared by dissolving 10.00gram of LiCN in water up to 500. mL? Hydrocyanic acid has a Ka-4.9x10^-10.

> So far I get molarity of 0.607M but I am not sure about the overall equation to set up an ice table

Answer 1

I don't believe you really need an ICE table to solve this. A simplified approach is as follows:

CN^- + H2O ==> HCN + OH^-

Since CN^- is a base, we can write Kb = [HCN][OH^-]/[CN^-]

[CN^-] = 10g LiCN/32.96 g/mol = 0.3034 moles/0.5 L = 0.607 M

Kb = 1x10^-14/4.9x10^-10 = 2.04x10^-5

2.04x10^-5 = (x)(x)/0.607 - x and assuming x is small relative to 0.607, we can neglect it

(2.04x10^-5)(0.607) = x^2 = 1.24x10^-5

x = 3.5x10^-3 = [OH-]

pOH = -log 3.5x10^-3 = 2.45

pH = 14 - 2.45 = 11.5

Answer 2

Hello Allie

Good work on calculating the molarity of the LiCN solution!

To set up the ICE table, we first need to write the reaction of LiCN with water. In this reaction, the Lithium ion (Li+) acts as a spectator ion, therefore, it is omitted from the reaction. The cyanide ion (CN-) is the weak base that reacts with water to form HCN and OH-.

R: CN- + H2O <==> HCN + OH-

I: 0.607 M 0 0

C: -x +x +x

E: 0.607 - x x x

(Allie: Wyzant's new update to this forum is messing up the formatting for the above ICE table. Therefore, I am going to re-write the values from the ICE table below, which I hope will be clearer to you.)

Initial: [CN-] = 0.607 M, [HCN] = 0, [OH-] = 0

Change: [CN-] = -x, [HCN] = +x, [OH-] = +x

Equil.: [CN-] = 0.607 - x, [HCN] = x, [OH-] = x

Since CN- is a weak base, we need to write its "Kb" expression.

Kb = [HCN] [OH-] / [CN-]

The Ka of HCN is given. Kb is calculated using the equation,

Kw = Ka x Kb

Kb = Kw / Ka = 1.0 x 10^-14 / 4.9 x 10^-10 = 2.0 x 10^-5

Now substitute the equilibrium concentrations of CN-, HCN and OH- into the Kb expression, you get

Kb = (x) (x) / (0.607 - x) = 2.0 x 10^-5

Since Kb is a small #, we can assume that the value of x is negligible when compared to the concentration of CN- (0.607 M). The Kb expression now simplifies to:

Kb = x^2 / 0.607 = 2.0 x 10^-5

x^2 = 0.607 x 2.0 x 10^-5

x^2 = 1.2 x 10^-5

x = square root (1.2 x 10^-5) = 3.5 x 10^-3

x = [OH-] = 3.5 x 10^-3 M

pOH = - log [OH-] = - log (3.5 x 10^-3) = 2.46

pH = 14.00 - 2.46 = 11.54