Question

based on the measured pH values, write the net ionic equation for the following
0.1M Na2CO3 = 10.85pH
0.1 M NaC2H3O2= 6.92pH
0.1M NH4NO3= 5.87pH
0.1M Cu(NO3)2= 3.23pH

Answer 1

(a): 0.1M Na2CO3 = 10.85pH:

Since pH is greater than 7, the solution is basic. The basic nature of the solution is due to hydrolysis of CO₃^2- to form H2CO3 and OH-. The net ionic equation is

CO₃^2-(aq) + 2H2O ---- > H2CO3 + 2OH^-(aq)

(b):0.1 M NaC2H3O2= 6.92pH:

CH3COO- undergoes hydrolysis to form CH3COOH. Now CH3COO- and CH3COOH act as buffer solution whose pH is 6.92. The net ionic equation for the hydrolysis reaction is

CH3COO-(aq) + H2O ----- > CH3COOH + OH-(aq)

(c): 0.1M NH4NO3= 5.87pH:

The dissociation of NH4+ occurs to give weak base NH3 and H+(aq). Due to the formation of H+ the solution becomes acidic.

The net ionic equation is

NH4+(aq) ----- > NH3 + H+(aq)

(d): 0.1M Cu(NO3)2= 3.23pH:

Cu2+ underges hydrolysis to form weak base Cu(OH)2 and strong acid HNO3. This makes the solution highly acidic.The net ionic reaction is

Cu2+(aq) + H2O ---- > Cu(OH)2(s) + 2H+(aq)