Astudent 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

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 HCl0.007874 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!

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