Question

# Consider the titration of 20.00 mL of 0.306M NH3 Calculate the pH of the solution after...

Consider the titration of 20.00 mL of 0.306M NH3 Calculate the pH of the solution after following volume of 0.460M HCl have been added. For NH3, Kb=1.8

NH3 + HCL ------------ NH4+ + Cl- Neutralization reaction

NH3 + H2O ====== NH4+ + OH- Equilibrium reaction of NH3

M= mol HCl/ V(L) ===== molHCl = M*V= 0,460M*0,010L= 0,046mol HCl

Inicially we have the molarity and volume of NH3 so we know how many mol we have to neutralize, we can calculate the mols of NH3 just like we just did with HCl.

molNH3= 0,360M*0,020L= 0,00612mol NH3

So we add an amount of HCl that will react 1:1 with NH3, so we can calculate the amount of NH3 that didn't react after we add the HCl, because is in excess.

molNH3 left= mol NH3 inicials - mol NH3 neutraliced = 0,00612mol-0,0046mol=0,00152molNH3

Because we have an amount of NH3 that wasn't neutraliced, we have the equilibrium reaction of the base, so we calculate the amount of OH- that we have as usual

NH3 + H2O ===== NH4+ + OH-

Inicially 0,00152 mol 0 0

equilibrium 0,00152-X X X

Kb= [NH4+][OH-]/[NH3] = X2/ (0,0152-X)

We asume that 0,00152-X is aproximatly the same as 0,00152, so we substitute

Kb= X2/ 0,00152 ----------- X= (Kb*0,00152)1/2 = (1,8*10-5*0,00152)=1,65*10-4M

pOH= -log(1,65*10-4)= 3,78

pH= 14-pOH = 10,22.

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