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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

Respuesta :

Willchemistry
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!
M(acid) x V(acid) = M(base) x V(base) 

You know all of those except the Molarity of the acid. So rearrange so as to solve for M(acid) 

M(acid) = M(base) x V(base)/V(acid) 

Plug in the values that you know, and solve: 

M(acid) = 0.227 

0.227M HCl