Formula for pH of a mixture of acid (3 dissociation steps) & base (1 dissociation step) :
(C_(a ) (K_a1 [H]^2 + 2K_a2 [H] + 3K_a3) / ([H]^3 + K_a1 [H]^2 + K_a2 [H] + K_a3 ) + K_w/[H] =
(C_(b ) (K_b1 [H] ) / (K_w+ [H] K_b1 ) + [H]
Calculating the pH of a mixture of phosphoric acid (with Ca=0.01, 3 dissociation steps) &
ammonia (with Cb=0.02, 1 dissociation step):
K_a1=10^(-2.148)
K_a2=10^(-7.199)
K_a3=10^(-12.35)
K_w= 10^(-14)
K_b1=10^(-4.75)
Left-hand side works out to be:
(7.11214×10^(-5) [H]^2+1.26482×10^(-9) [H]+1.34005×10^(-14)) / ([H]^3+ 7.11214×10^(-3) [H]^2+6.32412×10^(-8) [H]+4.46684×10^(-13) )+ 10^(-14)/([H])
Right-hand side works out to be:
(3.55656×10^(-7) [H])/(10^(-14)+1.77828×10^(-5) [H] )+ [H]
When I solve for [H], then I find:
[H] = 7.94328×10^(-6), with
LHS = RHS = 1.99×10^(-2)
So this would mean that pH = 5.10
However, Chembuddy pH calculator’s solution gives pH= 8.06
I checked this several ways, over and over again, and can’t find where I go wrong, I would very much appreciate if
someone could indicate what’s going on here, thanks a lot in advance!