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35.0 mL of acid with an unknown concentration is titrated with 24.6 mL of 0.432 M base
What is the concentration of the acid? Assume the acid contributes 1 mole of (H+) ions/mole of acid and the base contributes 1 mole of (OH-) ions/mole of base.

3.29 M
0.615 M
0.304 M

Respuesta :

meerkat18
We are given with the titration volume and concentration of the titrant  which is 24.6 mL of 0.432 M base. Given there is a one is to one ratio between the acid and the base, the equation becomes 26.4 ml (0.432 M) = 35 ml *(M) where M is the molarity of the acid. The concentration is 0.304 M.

Answer : The correct option is, 0.304 M

Explanation :

Using neutralization law,

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]n_1[/tex] = number of moles of acid = 1 mole

[tex]n_2[/tex] = number of moles of base = 1 mole

[tex]M_1[/tex] = concentration of acid= ?

[tex]M_2[/tex] = concentration of base = 0.432 M

[tex]V_1[/tex] = volume of acid = 35 ml

[tex]V_2[/tex] = volume of base = 24.6 ml

Now put all the given values in the above law, we get the concentration of the acid.

[tex]1mole\times M_1\times 35ml=1mole\times 0.432M\times 24.6ml[/tex]

[tex]M_1=0.304M[/tex]

Therefore, the concentration of the acid is, 0.304 M