Answer to Question #237023 in Physical Chemistry for Chicarizo

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

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)NaCNO(aq)+H2​O→HCNO(aq)+NaOH(aq) ....(1)....(1)

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

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

H^+ + H_2O\to H_3O^+H++H2​O→H3​O+ .....(4).....(4)

Hence , the concentration of CNO^- ,H^+CNO−,H+ and H_3O^+H3​O+ will be similar to concentration of HCNOHCNO

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

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

As per equation (1),(1), concentration of HCNOHCNO and NaOHNaOH will be similar to NaCNONaCNO with water in excess and total volume constant =0.20M=0.20M

So,[H_3O^+]=[CNO^-]=[OH]^-=[HCNO]=0.2M[H3​O+]=[CNO−]=[OH]−=[HCNO]=0.2M

pHpH of solution =-log[H^+]=-log(0.2)=0.7=−log[H+]=−log(0.2)=0.7