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A solution of HCl with a volume of 25.00 mL is titrated to the endpoint, with 0.250 M
NaOH. If it takes 34.56 mL of NaOH, what is the original concentration of HCl in the
solution?
HCl(aq) + NaOH(aq) → H20(l)+ NaCl(aq)

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Answer:

[tex]0.3456\ \text{M}[/tex]

Explanation:

[tex]V_1[/tex] = Volume of NaOH = 34.56 mL

[tex]V_2[/tex] = Volume of HCl = 25 mL

[tex]M_1[/tex] = Concentration of NaOH = 0.25 M

[tex]M_2[/tex] = Concentration of HCl

When endpoint is reached the number of moles of NaOH will be equal to the number of moles of HCl

[tex]M_1V_1=M_2V_2\\\Rightarrow M_2=\dfrac{M_1V_1}{V_2}\\\Rightarrow M_2=\dfrac{0.25\times 34.56}{25}\\\Rightarrow M_2=0.3456\ \text{M}[/tex]

Concentration of HCl is [tex]0.3456\ \text{M}[/tex].

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