Answer :

Final answer:

To titrate 6.00 g of KOH with 0.225M HCl, approximately 475.1 mL of HCl is needed. This calculation is based on the molar mass of KOH, the 1:1 reaction ratio, and conversion from moles to volume.

Explanation:

To determine how many milliliters of 0.225M HCl would be needed to titrate 6.00 g of KOH, we first need to convert the mass of KOH to moles using its molar mass. The reaction between HCl and KOH is a neutralization reaction where HCl reacts with KOH in a 1:1 mole ratio, as seen in the equation: HCl + KOH ightarrow KCl + H₂O.

First, calculate the number of moles of KOH:

Molar mass of KOH = 39.1 (K) + 16.0 (O) + 1.0 (H) = 56.1 g/mol

Moles of KOH = (mass of KOH) / (molar mass of KOH) = 6.00 g / 56.1 g/mol ≈ 0.1069 mol

Since the reaction ratio between HCl and KOH is 1:1, we need the same number of moles of HCl to react with the KOH:

Volume of HCl needed = (moles of KOH) / (molarity of HCl) = 0.1069 mol / 0.225 M ≈ 0.4751 L

Finally, convert the volume from liters to milliliters:

Volume of HCl needed in mL = 0.4751 L × 1000 mL/L ≈ 475.1 mL

Therefore, approximately 475.1 mL of 0.225M HCl is needed to titrate 6.00 g of KOH.

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Rewritten by : Barada