Question

A solution of sodium cyanate, NaCNO(aq), is prepared at a concentration of 0.20 M. Calculate the equilibrium concentrations of OH(aq), HCNO (aq), CNO(aq), and H3O+(aq) and the pH of the solution at 25°C.

Answer 1

Assuming water to be in excess and total volume constant, Sodium cyanate dissociates as follows:

$NaCNO(aq) + H_2O → HCNO (aq)+ NaOH (aq)$ $....(1)$

$HCNO\to H^++CNO^-$ $.......(2)$

$NaOH\to OH^- + Na^+$ $.........(3)$

$H^+ + H_2O\to H_3O^+$ $.....(4)$

Hence , the concentration of $CNO^- ,H^+$ and $H_3O^+$ will be similar to concentration of $HCNO$

And, concentration of of $OH^-$ will be similar to $NaOH.$

So, concentration of $NaCNO(aq)=0.20M$

As per equation $(1),$ concentration of $HCNO$ and $NaOH$ will be similar to $NaCNO$ with water in excess and total volume constant $=0.20M$

So,$[H_3O^+]=[CNO^-]=[OH]^-=[HCNO]=0.2M$

$pH$ of solution $=-log[H^+]=-log(0.2)=0.7$