Question

Calculate the pH during the titration of 25.00 mL of 0.1000 M LiOH(aq) with 0.1000 M HI(aq) after 24.1 mL of the acid have been added.

Answer 1

Number of moles of LiOH,n =molarity x volume in L

n= 0.1000M x 25.00 mL x10^-3 L/mL

= 25x10^-4 mol

Number of moles of HI added,n'=0.1000Mx24.1 mLx10^-3 L/mL

n'=24.1x10^-4 mol

HI + LiOH ----->LiI + H2O

So 1 mol LiOH reacts with 1 mol HI

24.1x10^-3 mol LiOH reacts with 24.1x10^-3 mol HI

So 25.0x10^-3 - 24.1x10^-3 = 0.9x10^-3 mol LiOH left unreacted

Which causes basic nature to the solution

Molaity of left base = molarity of [OH]= number of moles / total volume

= 0.9x10^-3 mol /(25.0+24.1)mL x10^-3 L/mL

= 0.0183 M

pOH= -log[OH-]

= - log 0.0183

= 1.74

pH= 14 -pOH

= 14 -1.74

= 12.26