Q: a 25.5 ml aliquot of hcl (aq) of unknown concentration was titrated with 0.113 m naoh (aq). it took 51.2 ml of the base to reach the endpoint of the titration. the concentration (m) of the acid was . a) 0.227. b) 1.02
c) 0.114
d) 0.454
e) 0.113

M ( HCl ) = ?

V ( HCl ) = 25.5 mL in liters : 25.5 / 1000 => 0.0255 L

M ( NaOH ) = 0.113 M

V ( NaOH ) = 51.2 mL / 1000 => 0.0512 L

number of moles NaOH:

n = M x V

n = 0.113 x  0.0512 => 0.0057856 moles of NaOH

mole ratio:

HCl + NaOH = NaCl + H2O

1 mole HCl 1 mole NaOH
( moles HCl )   0.0057856 moles NaOH

(moles HCl ) =  0.0057856 x 1 / 1

=  0.0057856 moles of HCl

M ( HCl ) = n / V

M =  0.0057856 / 0.0255

= 0.227 M

Answer A

hope this helps!

Bis the