Question

if you could show work to all parts of question number 4, a-g

Answer 1

0.53 M means 0.53 mole in 1000 ml but he has transfered only 25 ml

So 25ml 0.53 M HF contains(0.53/1000)*25 = 0.01325mole HF

a) Number of mole of weak acid = 0.01325 mole

b) Since 0.01325 mole acid can be neutralise by 0.01325 mole base

Number of mole of base to neutralise = 0.01325 mole

c) 0.096M of NaOH means 0.096 mole in 1000ml

so for 1 mole we have to take 1000/0.096 ml

but we need 0.01325 mole , hence we have to take (1000/0.096) * 0.01325 = 138ml

d) 1/2 volume of 138 is 69 ml

in 69 ml there will be 0.01325/2 = 0.0066mole NaOH

at a section 4 questions will be answered , not more than that