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 re
ach 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
2 answers:
M(acid) x V(acid) = M(base) x V(base)
<span>You know all of those except the Molarity of the acid. So rearrange so as to solve for M(acid) </span>
<span>M(acid) = M(base) x V(base)/V(acid) </span>
<span>Plug in the values that you know, and solve: </span>
<span>M(acid) = 0.227 </span>
<span>0.227M HCl </span>
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 <span> 0.0512 => 0.0057856 moles of NaOH
mole ratio:
</span><span>HCl + NaOH = NaCl + H2O
</span><span>
1 mole HCl -------------- 1 mole NaOH
( moles HCl ) ----------- </span><span> 0.0057856 moles NaOH
</span>
(moles HCl ) = <span> 0.0057856 x 1 / 1
</span>
= <span> 0.0057856 moles of HCl
</span>
M ( HCl ) = n / V
M = 0.0057856 / <span>0.0255
</span>
= 0.227 M
Answer A
hope this helps!
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