cutegirl1000
contestada

Please Help!!
A researcher takes a 10 mL sample of HCl from a bottle and titrates it with NaOH. It was found that 22.4 mL of 0.25 M NaOH were required to reach the equivalence point. What is the concentration of the HCl in original bottle?

Respuesta :

Answer : The concentration of the HCl in original bottle is 0.56 M.

Explanation :

According to neutralization law:

[tex]M_1V_1=M_2V_2[/tex]

where,

[tex]M_1[/tex] = concentration of  HCl

[tex]M_2[/tex] = concentration of NaOH

[tex]V_1[/tex] = volume of HCl

[tex]V_2[/tex] = volume of NaOH

Given:

[tex]M_1[/tex] = ?

[tex]M_2[/tex] = 0.25 M

[tex]V_1[/tex] = 10 mL

[tex]V_2[/tex] = 22.4 mL

Now put all the given values in the above formula, we get:

[tex]M_1\times 10mL=0.25M\times 22.4mL[/tex]

[tex]M_1=0.56M[/tex]

Therefore, the concentration of the HCl in original bottle is 0.56 M.

Otras preguntas