NX3(aq) + H2O (l) <---> HNX3+ (Aq)+OH-(aq) Kb=[HNX3+][OH-]/[NX3] Part a: If Kb f

Question

NX3(aq) + H2O (l) <---> HNX3+ (Aq)+OH-(aq) Kb=[HNX3+][OH-]/[NX3] Part a: If Kb for NX3 is 1.5*10^-6, what is the pOH of a .175M aqueous solution of NX3 Part b: If Kb for NX3 is 1.5*10^-6, what is percent inonization of a .325M aqueous solution of NX3 Part c: If Kb for NX3 is 1.5*10^-6, what is the pka for the following reaction? HNX3+(aq) + H2O(l) <---> NX3 (aq) + H3O+ (aq) NX3(aq) + H2O (l) <---> HNX3+ (Aq)+OH-(aq) Kb=[HNX3+][OH-]/[NX3] Part a: If Kb for NX3 is 1.5*10^-6, what is the pOH of a .175M aqueous solution of NX3 Part b: If Kb for NX3 is 1.5*10^-6, what is percent inonization of a .325M aqueous solution of NX3 Part c: If Kb for NX3 is 1.5*10^-6, what is the pka for the following reaction? HNX3+(aq) + H2O(l) <---> NX3 (aq) + H3O+ (aq) If Kb for NX3 is 1.5*10^-6, what is percent inonization of a .325M aqueous solution of NX3 Part c: If Kb for NX3 is 1.5*10^-6, what is the pka for the following reaction? HNX3+(aq) + H2O(l) <---> NX3 (aq) + H3O+ (aq)

Explanation / Answer

Kb=[HNX3+][OH-]/[NX3]

Kb for NX3 is 1.5*10^-6

initial conc of NX3, HNX3,OH- are 0.175 ,x,x,

final conc are 0.175-x,x,x

kb=x2/0.175-x

1.5 * 10-6=x2/0.175-x

x=0.512 x 10-3 M

pOH=-log(OH-)

=-logx

=-log ( 0.512 * 10-3)

pOH =3.29

Part B

Kb=x2/0.325-x

1.5 * 10-6=x2/0.325-x

x=0.70 x 10-3 M

percentage ionization = ( 0.0007/0.325) x 100

= 0.215 %

Given Kb=1.5 x 10-6

we know that Ka=Kw/kb

=10-14/(1,5 x 10-6)

=6.67 x 10-9

pKa=-logka

=-log ( 6.67 x 10-9)

pKa=8.17

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