Question

A student neutralized 16.4 milliliters of HCl by adding 12.7 milliliters of 0.620 M KOH. What was the molarity of the HCl acid?

(1) 0.168 M (3) 0.620 M
(2) 0.480 M (4) 0.801 M

Answer 1

V ( HCl ) = 16.4 mL / 1000 => 0.0164 L

M( HCl) = ?

V( KOH) = 12.7 mL / 1000 => 0.0127 L

M(KOH) = 0.620 M

Number of moles KOH:

n = M x V

n = 0.620 x 0.0127

n = 0.007874 moles of KOH

number of moles HCl :

HCl + KOH = H2O + KCl

1 mole HCl ------ 1 mole KOH
? mole HCl--------0.007874 moles KOH

moles HCl = 0.007874 * 1 / 1

= 0.007874 moles of HCl

M = n / V

M = 0.007874 / 0.0164

= 0.480 M

Answer (2)

hope this helps!