How many mL of 0.225M HCl would be required to titrate 6.00 grams of KOH?
See above.
To solve this question, you will need to use the concept of stoichiometry and the balanced chemical equation between HCl (Hydrochloric acid) and KOH (Potassium hydroxide).
The balanced equation for the reaction between HCl and KOH is:
HCl + KOH -> KCl + H2O
From the balanced equation, you can see that the mole ratio between HCl and KOH is 1:1. This means that for every 1 mole of HCl, you will need 1 mole of KOH to react completely.
Now, let's calculate the number of moles of KOH in 6.00 grams:
Step 1: Convert grams of KOH to moles using the molar mass of KOH.
Molar mass of KOH = 39.10 g/mol (K) + 16.00 g/mol (O) + 1.01 g/mol (H) = 56.11 g/mol
Moles of KOH = mass of KOH / molar mass of KOH
= 6.00 g / 56.11 g/mol
Step 2: Determine the moles of HCl required. Since the mole ratio between HCl and KOH is 1:1, the moles of HCl required will be the same as the moles of KOH.
Moles of HCl = Moles of KOH
Now, let's calculate the volume of 0.225M HCl required to react with the calculated moles of HCl:
Step 3: Use the formula for molarity: Molarity (M) = Moles (mol) / Volume (L)
Rearranging the formula, Volume (L) = Moles (mol) / Molarity (M)
Volume of HCl = Moles of HCl / Molarity of HCl
= Moles of HCl / 0.225 M
Finally, you can calculate the volume of 0.225M HCl required in milliliters (mL) by converting the volume from liters to milliliters:
Volume of HCl (mL) = Volume of HCl (L) x 1000
By following these steps and performing the necessary calculations, you should be able to calculate the required volume of 0.225M HCl to titrate 6.00 grams of KOH.