Question

Suppose you believe your unknown compound is NaOH, and you prepare 20 mL of a 0.1 mol/L solution of it. The stockroom provides a 0.2 mol/L HCl solution as the titrant. What volume of HCl (in mL) is needed to reach the equivalence (stoichiometric) point if your unknown compound is NaOH? NaOH(aq) + HCl(aq) --> NaCl(aq) + H2O(l)

Answer 1

Answer : Volume of HCl ( in ml) is 10 ml.

Solution : Given, Concentration of NaOH = 0.1 mol/L

Volume of NaOH = 20 ml

Concentration of HCl = 0.2 mol/L

Volume of HCl = ?

Formula used : Moles = Volume × Concentration

In the reaction, we see that 1 mole of NaOH react with the 1 mole of HCl.

so,

Moles of NaOH = Moles of HCl

Moles of HCl = Moles of NaOH = Volume of NaOH ( in L) × Concentration of NaOH

Converstion ml into L : 1000 ml = 1 L

Moles of HCl =
`(20)/(1000)` × 0.1

= 0.002 mol

Volume of HCl = Moles / Comcentration of HCl

=
`(0.002 mole)/(0.2mole/L)`

= 0.01 L

Volume of HCl (in ml) = 0.01 × 1000 = 10 ml