Question

i mix 500ml of 0.250 M HCNO with 250ml of 0.333 M NaCNO. i then add 25ml of 1.00 M NaOH. A) how many moles of CNO- are in the beaker before the reaction takes place. B) what is the limiting reactant for the neutralization reaction

Answer 1

A )

The question relates to ' common ion effect ' ( ie. the degree of dissociation of

a soluble electrolyte is supressed due to addition of a common ion ' ) . HCNO is

a weak acid with K_a = 3.5 x 10^-4 hence its degree of dissociation ( ie. number of

moles of the acid dissociated out of one mole ) is calculated using Ostwald's dilution

law. Accordingly

= SQRT ( K_a / C ) , where represents degree of dissociation of the acid

..............& C = concentration in moles

calculated* as =( 0.250 x 500 ) / 750

....................... = 0.166 M

.......... So , = SQRT {( 3.5 x 10^-4 ) / 0.166 }

........................= 0.0459 x 10^-2

...................or, = 4.59 x 10^-4 M

This represents concentrations of H⁺  ion & also NCO^- ion yielded by ionization of

HCNO in solution , which is much much less than unity ( <<< 1.0 ) and is therefore

insignificant. However , the total [ NCO^- ] = [ NCO^- ]  produced by ionization of the salt

NaCNO + [ NCO^- ]yielded by ionization of HCNO

NaCNO is a strong electrolyte and therefore gets completely ionized

The [ NaCNO] in 750 ml of solution is = (0.333 x 250 ) / 750

..............................................................= 0.111 M

therefore[ NCO^- ]   ion produced...........= ( 0.111 + 4.59 x 10^-4 ) / L

again since( 4.59 x 10^-4 ) is <<< 1.0 , it can be ignored

& the total NCO^- in 750 ml of solution = ( 0.111 x 750 ) /1000

................................................................= 8.325 x 10^-2 moles  

B)

H⁺ has to partcipate in neutralization reaction with a base its concentration decides the

limiting reagent ,and a limiting reagent is one which gets consumed first, Therefore on addition of 25 ml of 1M NaOH solution HCNO should be considered as the limiting reagent.

...............................................................................................................................................

* strength or concentrations of HCNO ,& NaCNO are calculated using relation,

M1V1 = M2V2