Question

Titration: Can someone help me understand this concept. Sample question below.

If 10.0mL of 0.250 mol/L NaOH(aq) is added to 30.0mL of 0.17 mol/L HOCN(aq), what is the pH of the resulting solution?

answer is 3.74 [I need explanation how to get it]

Answer 1

start with a balanced net ionic equation.

Answer 2

NaOH + HOCN -> H2O + NaOCN?

Answer 3

Net ionic... so you can omit the Na...since it is the same on both sides of the equation

Answer 4

oops sorry! my bad.. should be \(\sf OH^- + H^+ -> H_2O_{(l)}\)? because I think Na and OCN will be spectator ions..

Answer 5

why do i need to get the net ionic equation though?

Answer 6

cyanic acid is a weak acid ..
so the overall equation is between HOCN + OH-

Answer 7

go ahead and complete the reaction

Answer 8

\(\sf OH^- + HOCN -> H_2O_{(l)} + OCN^- ? \)

Answer 9

ok great... so that is the net ionic equation....now you have to determine the equilibrium constant for this equation

Answer 10

okay.. wait, i'm really confused..
equilibrium constant is like Keq= [OCN-][H2O]/ [OH][HOCN] right?
so what would be the concentration of H2O and the OCN- ?

Answer 11

you would not include H20 , you only include aqueous and gas phase material.
You will be calculating the [OCN-] on the way to getting the pH

Answer 12

never include pure liquids or solids in equilibrium expressions

Answer 13

oh yeah, I forgot about that.
how to calculate [OCN-] if I don't know Keq? should I use Ka or Kb? Ka right?

Answer 14

hmm i think something's wrong...