36.38 ml of a 0.167 M potassium hydroxide solution is required to neutralize 16.6 mL of a 0.366 M nitric acid solution.
Given that 16.6 mL of a 0.366 M nitric acid solution is neutralized by 0.167 M potassium hydroxide solution.
The balanced equation for the reaction is:
HNO₃ + KOH → KNO₃ + H₂O
From the equation given above,
Mole ratio of the acid, HNO₃ (nA) = 1
Mole ratio of the base, KOH (nB) = 1
From the question:
Molarity of the acid, HNO₃ (Ma) = 0.366 M
Volume of the base, KOH (Vb) =
Molarity of the base, KOH (Mb) = 0.167 M
Volume of the acid, HNO₃ (Va) = 16.6 ml
Now,
MaVa/ MbVb = nA/nB
0.366 × 16.6 / 0.167 × Vb = 1/1
6.0756 = 0.167 × Vb
Vb = 36.38 ml
Therefore, 36.38 ml of potassium hydroxide solution is required for the reaction.
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