What volume (mL) of 0.250 M HNO3 is required to titrate (neutralize) a solution containing 0.276 g of KOH?

Volume = 19.68 ml

Explanation:

The equation of the reaction is given as;

HNO3 + KOH ---> KNO3 + H2O

1 mol of HNO3 reacts 1 mol of KOH

Converting 0.276 g of KOH to mol;

Number of moles = Mass / Molar mass

Number of moles = 0.276g / 56.1056 g/mol

Number of moles = 0.00492

Since the mole relationship is 1 = 1;

This means 0.00492 mol of HNO3 reacts with 0.00492 mol of KOH

The relationship between molarity and volume id given as;

Molarity = Number of moles / Volume

Volume = Number of moles / Molarity = 0.00492 mol / 0.250 M

Volume = 0.01968 L

Volume = 19.68 ml

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explanation:

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