Question

`50ml` of solution which is `0.10M` in the acid `"HA"`, `pKa=4.00`, and `0.050M` in `"HB"`, `pKa=8.00`, is titrated with `0.20M` `"NaOH"`. How can I calculate the pH at (i) the first eq.point (ii) at the second eq.point.? answer is (i) 6.15; (ii) 10.23

Answer 1

pH 6 and pH = 10.23

Explanation:

At the first equivalent point you have the reaction:
`HA+ NaOH = NaA + H_2O`
that you obtain consuming 25 mL of NaOH
In the solution we have HB whose concentration is now:
`C_bf = C_bi xx(V_i)/(V_f) = 0.050 M xx (50 mL)/((50+25) mL)=0.0667(mol)/L`
and NaA whose concentration is `M=n_s/V_t= (M_a xx V_a)/(V_a+V_b)= (50mL xx 0,10(mol)/(L))/((50+25)mL)=(5mmol)/(75mL)= 0.0667M`

Hence the concentration are the same you can use the formula of the pH of a biprotic acid after the titration of its first `H^+` whwere the concentration doesn't influence the pH because you have at the same tima an acid dissociation and an hydrolisis basic reaction
pH =1/2( pk_1+pK_2)= 1/2 (4+8) = 6.00
pharaps there are better formulas to obtain the result that you have

for the second equivalent point you have the reaction:
`HB+ NaOH = NaB + H_2O`
the resulting pH is given by the stronger basic hydrolisis of NaB whose concentration is `M=n_s/V_t= (M_b xx V_b)/(V_a+V_b1+V_b2)= (50mL xx 0.05(mol)/L)/((50+25+12,5)mL)= (2.5mmol)/(87.5mL)= 0.0286M`
`[OH^-]=sqrt(K_w/K_b xx Cs)= sqrt(10^(-14)/10^(-8) xx0.0286)=0,169 xx 10^(-3)(mol)/L`
pOH = 3,77
pH= 14-3,77= 10,23