How many milliliters of 0.326 M HCl would be required to titrate 5.48 g KOH?
To determine the number of milliliters of 0.326 M HCl required to titrate 5.48 g of KOH, we need to use the concept of stoichiometry and the equation of the reaction between the two compounds.
The balanced equation for the reaction between HCl and KOH is:
HCl + KOH → KCl + H2O
From this balanced equation, we can see that the stoichiometric ratio between HCl and KOH is 1:1. This means that one mole of HCl reacts with one mole of KOH.
First, we need to calculate the number of moles of KOH using its given mass and molar mass.
Molar Mass of KOH:
K (potassium) = 39.10 g/mol
O (oxygen) = 16.00 g/mol
H (hydrogen) = 1.01 g/mol
Molar Mass of KOH = K + O + H = 39.10 + 16.00 + 1.01 = 56.11 g/mol
Number of moles of KOH = Mass / Molar Mass
Number of moles of KOH = 5.48 g / 56.11 g/mol ≈ 0.098 moles
Since the stoichiometric ratio between HCl and KOH is 1:1, we can conclude that we need 0.098 moles of HCl to react with 0.098 moles of KOH.
Now, we can use the concentration of the HCl solution to determine the volume needed.
Concentration of HCl = 0.326 mol/L
Number of moles of HCl = 0.098 moles
Volume of HCl solution = Number of moles / Concentration
Volume of HCl solution = 0.098 moles / 0.326 mol/L ≈ 0.301 L
Since the question asks for the volume in milliliters, we can convert the volume from liters to milliliters.
Volume of HCl solution in milliliters = 0.301 L × 1000 mL/L = 301 mL
Therefore, approximately 301 milliliters of 0.326 M HCl would be required to titrate 5.48 g of KOH.
HCl + KOH ==> KCl + H2L
mols KOH = grams/molar mass
mols KOH = mols HCl from the 1:1 ratio (coefficients) in the balanced equation.
Then M HCl = mols HCl/L HCl. You know mols and M, solve for L and convert to mL.