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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

Verified Answer :

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,

\(n_1M_1V_1=n_2M_2V_2\)

where,

\(n_1\) = number of moles of acid = 1 mole

\(n_2\) = number of moles of base = 1 mole

\(M_1\) = concentration of acid= ?

\(M_2\) = concentration of base = 0.432 M

\(V_1\) = volume of acid = 35 ml

\(V_2\) = volume of base = 24.6 ml

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

\(1mole\times M_1\times 35ml=1mole\times 0.432M\times 24.6ml\)

\(M_1=0.304M\)

Therefore, the concentration of the acid is, 0.304 M

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