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Calculate the volume of 0.882 M HNO\(_3\) that can be neutralized by 55.3 mL of a 1.70 M KOH solution.

Answer :

The volume of the 0.882 M HNO3 solution that can be neutralized by 55.3 mL of a 1.70 M KOH solution is approximately 106.94 mL.

the volume of 0.882 M HNO3 that can be neutralized by 55.3 mL of a 1.70 M KOH solution, we can use the concept of stoichiometry.


First, let's write the balanced chemical equation for the neutralization reaction between HNO3 and KOH:

HNO3 + KOH -> KNO3 + H2O

From the balanced equation, we can see that 1 mole of HNO3 reacts with 1 mole of KOH to produce 1 mole of KNO3 and 1 mole of water.

Now, let's calculate the moles of KOH present in the 55.3 mL solution:

moles of KOH = volume (in L) x concentration (in mol/L)
= 55.3 mL x 1.70 mol/L
= 94.21 mmol (millimoles)

Since the reaction is 1:1 between HNO3 and KOH, the moles of HNO3 required to neutralize the KOH will also be 94.21 mmol.

Now, let's calculate the volume of the 0.882 M HNO3 solution that contains 94.21 mmol of HNO3:

volume of HNO3 = moles of HNO3 / concentration of HNO3
= 94.21 mmol / 0.882 mol/L
= 106.94 mL

Hence, the volume of the 0.882 M HNO3 solution that can be neutralized by 55.3 mL of a 1.70 M KOH solution is approximately 106.94 mL.

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