Using the example of [NaC2H3O2?]= 100.0mL of 1.10M
[H+] to be added with HCl= 2.0M @ 250.0mL
[OH-] added after HCl= 0.55M @ 500.0mL (NaOH?)
Step 1: NaC2H3O2? → Na+ (and) C2H3O2- {1 to 1 ratio} with all in 100.0mL, 1.10 Molarity. So 1.10 Molarity in 100.0mL is 0.110 Moles.((1000.0mL=1.0000L))
pH=pKa + (HA ÷ A-)
this will be used to calculate pH, while Ka(HC2H3O2)=1.8E5
Ka=1.8E5=( [H+]1 [C2H3O2-]1)
([HC2H3O2]1
[H+] [C2H3O2] [HC2H3O2]
Initial 0.00 Moles 0.11 Moles 0.00 Moles
Change(Δ) +0.50Moles-X -x +x
Equilibrium 0.50 Moles-X (0.110-X)moles X Moles
Initial Calculation will make primary equilibrium before added H+... 0.11Moles C2H3O2 → Na+ (and) C2H3O2- then the 1.8E5=Ka=H+ and A- (over) HA... [H+]0=0.00M [C2H3O2-]0=1.10M [Acetic Acid(HC2H3O2)]0=0.00M
since there will be a change in volume, moles are used. 0.11Moles C2H3O2 in 100.0mL
HCl to be added... 2.00 Molar at 250.0mL
HCl → (SA) H+ (and) Cl-
There is a volume change; 100.0mL initial has 250.0mL added to it., new volume = 350.0 mL...
Since it is {1to1} 2.00 Molar HCl is added. 0.50 Moles H+ are added;,
1.8E-5=(0.50-x)moles (times) (0.110-x)Moles (over/) X Moles
Through interactions of present molecules
The concentrations change/ Also the volume, so the remaining OH- or H+ from equilibrium of the WA or WB will be sole contributors to pH.