Question

Calculate the pH during the titration of 50.00 mL of 0.1000M phosphoric (H₃PO₄; K_a1=7.1x10^-3, K_a2=6.3x10^-8, K_a3=4.2x10^-13) after adding 28.2 mL of 0.1000M NaOH.

All volumes should have 2 decimal places

Answer 1

H3PO4+ 3NaOH ------ Na3PO4 + H2O

1 mole of H3PO4 neutralizes with 3 moles of NaOH. Molar ratio of reactants = 1:3

moles of H3PO4= molarity* volume = 0.1*50/1000 =0.005 moles of NaOH= 0.1*28.2/1000 =0.00282

actual ratio =0.005 :0.00282 = 1.8 :1

so excess is H3PO4. moles of H3PO4 remaining = 0.005-0.00282/3 = 0.0041

volume after mixing = 50+28.2= 78.2ml= 78.2/1000 = 0.0782 L

Concentration of H3PO4= 0.0041/0.0782= 0.0524

H3PO4 ionizes as

H3PO4 + H2O -------------> H2PO4- + H3O+ (1) H2PO4-2 + H2O-------> HPO4-2 + H3O+ (2) and HPO4-2+ H2O -----> PO4-3 + H3O+ (3)

let us consider each reaction

let x= drop in concentration of H3PO4 through reaction 1 to reach equilibrium

at equilibrium [H2PO4-] = [H3O+] =x and H3PO4= 0.0524-x

Ka1= x²/ (0.0524-x) =7.1*10^-3, when solved using excel, [H3O+] = [H2PO4-2] = 0.0161, [H3PO4]= 0.0524-0.0161= 0.0363

for second reaction similarly x²/(0.0161-x)= 6.3*10-8, looking at the magnitude of Ka2,

x²/0.0161= 6.3*10^-8, x= 3.18*10-5

thirs reaction is much slower as indicated by concentration in the second reaction

hence pH= -log[H3O+] =-log (0.0161)= 1.79