Question

How many milliliters of 0.250 M KoH will be needed to titrate 12.53 mL of 0.130 M `HNO_3`?

Answer 1

You will need 6.516 mL of KOH to reach the equivalence point.

Explanation:

Since KOH has only one `OH^-` ion in its formula, and `HNO_3` is monoprotic (donates only one hydrogen ion), the titration reaches an equivalence point when

Moles `HNO_3` = moles `KOH`

And since moles (of solute) = concentration x volume, we can write

`M_a*V_a=M_b*V_b`

(the `a` and `b` representing the acid and base, respectively)

We know the first three quantities in this equation. We find the fourth (`V_b`) as follows:

`(0.130)*(0.01253)=(0.250)V_b`

`V_b =((0.130)*(0.01253))/(0.250) = 0.006516 L`

or `6.516 mL`

Answer 2

Approx. `6-7*mL..........`

Explanation:

We need (i) a stoichiometric equation........

`HNO_3(aq) + KOH(aq) rarr KNO_3(aq) + H_2O(l)`

And then (ii) equivalent quantities of potassium hydroxide and nitric acid:

`"Moles of nitric acid"=12.53*mLxx10^-3L*mL^-1xx0.130*mol*L^-1=1.63xx10^-3mol`.

And thus we need `(1.63xx10^-3mol)/(0.250*mol*L^-1)xx10^3*mL*L^-1=6.52*mL`