Question

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

Answer 1

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* M_1* 35ml=1mole* 0.432M* 24.6ml`

`M_1=0.304M`

Therefore, the concentration of the acid is, 0.304 M

Answer 2

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.