How many milliliters of 0.250 M KoH will be needed to titrate 12.53 mL of 0.130 M HNO3?
Answers
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 HNO3 is monoprotic (donates only one hydrogen ion), the titration reaches an equivalence point when
Moles HNO3 = moles KOH
And since moles (of solute) = concentration x volume, we can write
Ma⋅Va=Mb⋅Vb
(the a and b representing the acid and base, respectively)
We know the first three quantities in this equation. We find the fourth (Vb) as follows:
(0.130)⋅(0.01253)=(0.250)Vb
Vb=(0.130)⋅(0.01253)0.250=0.006516L
or 6.516mL
or
Approx. 6−7⋅mL..........
Explanation:
We need (i) a stoichiometric equation........
HNO3(aq)+KOH(aq)→KNO3(aq)+H2O(l)
And then (ii) equivalent quantities of potassium hydroxide and nitric acid:
Moles of nitric acid=12.53⋅mL×10−3L⋅mL−1×0.130⋅mol⋅L−1=1.63×10−3mol.
And thus we need 1.63×10−3mol0.250⋅mol⋅L−1×103⋅mL⋅L−1
6.52⋅mL
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______________ ________________
Since KOH has only one OH- ion in its formula, and HNO3
is monoprotic (donates only one hydrogen ion), the titration reaches an equivalence point when Moles
HNO3= moles KOH
And since moles (of solute) = concentration x volume, we can write
Ma⋅Va=Mb⋅VL(the a and brepresenting the acid and base, respectively)
We know the first three quantities in this equation. We find the fourth (Vb) as follows:
(0.130)⋅(0.01253)=(0.250)VbVb=(0.1300.250=0.006516L or 6.516mL