10 mL of a 1 M HCl solution was added to a flask and titrated against a solution of 1.7 M NaOH.?

Question

How many mililiters is of NaOH was required to neutralize the acid?

Answer 1

(1.7 M)*x = (1M)(10 ml)
x = (1M)(10 ml)/(1.7 M)
x ≈ 5.88 ml

Answer 2

one hundred ml of 0.1N HCl grew to become into taken care of with 30 ml of of 0.2 N NaOH subsequently 60ml of HCl is ate up. now ultimate 40 ml of 0.1N HCl is taken care of with 0.25N KOH answer. word normality equation N1V1=N2V2 0.one million*40=0.25*V2 V2=0.one million*40/0.25 =16ml