Question

You used 15.56 mL of NaOH (0.1044 M) to neutralize 10.00 mL of HA. Knowing that Ka (HA) = 2.0 x 10^-5 and kb (A^-) = 5.0 x 10 ^-10. Calculate the pH at the beginning of the titration, before adding any NaOH to one decimal place.

Answer 1

From M1V1= M2V2

where M1=Molarity of NaOH and V1= volume of NaOH, M2= molarity of HA V2= volume of HA= 10ml

M2= 15.56*0.1044/10=0.1624M

this is the molarity of HA at the begining of titration

HA----> H+ A

Ka= [H+] [A-]/[HA]

let x= dissocition of HA

at Equilibrium [HA]= 0.1624-x [A-]= [H+]=x

hence x²/(0.1624-x)= 2*10^-5

tihs can be solved by either quadratic equation or assuming soome value of x and matching LHS and RHS which give x= 0.000569

pH= -log(0.000569)=3.24