1. Ka=([Acid-][H+])/[Acid]; 4.3e-7=([HCO3-][H3O+])/0.08; 3.44e-8=([HCO3-][H3O+]); Minus one H from one is plus one H to another, 1:1 ratio so [HCO3-]=[H3O+]=x. sqrt(3.44e-8)=x; x=0.00018547236. =1st dissociation. 2nd dissociation= 5.6e-11=([CO3(2-)][H+])/[HCO3-]; 5.6e-11=([CO3(2-)][H+])/0.00018547236; 1.0386452e-14=[CO3(2-)][H+]; sqrt(1.0386452e-14)=[CO3(2-)]=[H+]=1.01913944e-7; [H+]=0.00018547236+1.01913944e-7=0.00018557427; pH=-log[H+]; ph=-log[0.00018557427]=3.73148223917.

3. Kw=KbKa; Kw=[H+][OH-]; 1e-14=KaKb; 1e-14=Ka*1.76e-5; Ka=5.6818182e-10. NH3(aq)+H2O(aq)--->NH4(aq)

Kb=[products]/[reactants]; 1.76e-5=[OH][NH4]/0.188; 1.76e-5*0.188=[OH]^2; sqrt(0.0000033088)=[OH]=0.00181901072 : pOH=2.7401647415; pH=-log[H+]; pOH=-log[OH-]; [H+][OH-]=1e-14;
[H+]0.00181901072=1e-14; 1e-14/0.00181901072=H= 5.4974937e-12 : pH=11.2598352597
NH3+H2O:NH4+OH. Each ammonium is a hydroxide.
Determine the pOH of a 0.188 M NH3 solution at 25°C. The Kb of NH3 is 1.76 × 10-5.

hydronium ion concentration in an aqueous solution with a pOH of 4.33; -4.33=log[OH]; e^(-4.33)=[OH]= 0.01316754749; H*0.01316754749=1e-14; H=1e-14/0.01316754749=7.5944287e-13