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

2 Answers
Dwight
Apr 17, 2017

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

anor277
Apr 17, 2017

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

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