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If 5.25 mL of an unknown HCl sample was titrated with 25.3 mL of a 0.00100 M solution of NaOH to its equivalence point, what is the initial concentration of the acid?

4.64 × 10^-3 M


2.08 × 10^-2 M


4.82 × 10^-3 M


8.28 × 10^-4 M

Respuesta :

NaOH +HCl ---> NaCl +H2O

n (mol HCl) = n (mol  NaOH)
M- molarity
V - volume

M(HCl)V(HCl) = M(NaOH)V(NaOH)
M(HCl)= M(NaOH)V(NaOH)/V(HCl) 

M(HCl)= 10⁻³*25.3 ml/5.25 ml=4.82 × 10⁻³ M molarity HCl

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