Question

Please help me. Thanks

What is the pH of a solution prepared by dissolving 0.445g of Lithium cyanate, LCNO, in 500.0 mL of Water? What is the equilibrium concentration of hydrogen sulfite in a 0.150 M solution of H_2 SO_3

Answer 1

For the first question:

Let's write the ionic reaction:

CNO^- + H₂O <---------> HCNO + OH^-

Let's calculate the moles of CNO:

Molecular Weight = 6.94 + 16 + 14 + 12.01 = 48.95 g/mol

moles = 0.445 g / 48.95 g/mol = 0.0091 moles

[CNO] = 0.0091 moles / 0.5 L = 0.0182 M

The reported value of the Ka for HCNO is 3.54x10^-4 which means that the Kb is:

Kb = Kw/Ka = 1x10^-14 / 3.54x10^-4 = 2.82x10^-11

Let's write the reaction again and a ICE chart:

r: CNO^- + H₂O <---------> HCNO + OH^-

i: 0.0182 0 0

e. 0.0182-x x x

Kb = [HCNO][OH^-] / [CNO^-]

2.82x10^-11 = x² / 0.0182-x but Kb is really small so 0.0182-x can be approximated to 0.0182 so:

2.82x10^-11(0.0182) = x²

x = 7.16x10^-7 M = [OH^-]

pOH = -log(7.16x10^-7) = 6.14

Finally the pH:

pH = 14-6.14 = 7.86

For the second question you'll do exactly the same as above, but with the value of Ka for H₂SO₃. From there, solve for x, you'll have something similar to this.

Hope this helps