A volume of 80.0mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What was the molarity of the KOH solution if 21.7mL of 1.50 M H2SO4 was needed? The equation is
2KOH(aq)+H2SO4(aq)→K2SO4(aq)+2H2O(l)
2KOH(aq)+H2SO4(aq)→K2SO4(aq)+2H2O(l)
mols H2SO4 = M x L
mols KOH = 2*mols H2SO4
M KOH = mols KOH/L KOH
To find the molarity of the KOH solution, we need to use the balanced chemical equation and the concept of stoichiometry.
The balanced chemical equation shows that the mole ratio between KOH and H2SO4 is 2:1. This means that 2 moles of KOH react with 1 mole of H2SO4.
First, let's calculate the number of moles of H2SO4 used in the titration:
Moles of H2SO4 = (volume of H2SO4) x (molarity of H2SO4)
= 21.7 mL x 0.00150 mol/mL = 0.0326 mol
Since the mole ratio between KOH and H2SO4 is 2:1, the number of moles of KOH used in the titration is also 0.0326 mol.
Therefore, the number of moles of KOH in the 80.0 mL solution is also 0.0326 mol.
Now let's calculate the molarity of the KOH solution:
Molarity of KOH = (moles of KOH) / (volume of solution in liters)
= 0.0326 mol / 0.0800 L = 0.4075 M
Therefore, the molarity of the KOH solution is approximately 0.4075 M.
To determine the molarity of the KOH solution, we need to use the equation stoichiometry and the concept of molarity.
First, let's find the moles of H2SO4 used. We can use the given volume and molarity of H2SO4:
moles of H2SO4 = volume of H2SO4 (in L) x molarity of H2SO4
Given:
Volume of H2SO4 = 21.7 mL = 21.7/1000 L = 0.0217 L
Molarity of H2SO4 = 1.50 M
moles of H2SO4 = 0.0217 L x 1.50 M = 0.03255 moles
Next, using the balanced equation, we can determine the mole ratio between H2SO4 and KOH. For every 2 moles of KOH, there is 1 mole of H2SO4.
According to the balanced equation:
2 KOH(aq) + H2SO4(aq) → K2SO4(aq) + 2 H2O(l)
moles of KOH = 0.03255 moles H2SO4 x (2 moles KOH / 1 mole H2SO4) = 0.0651 moles KOH
Finally, we can calculate the molarity of KOH using the volume of the KOH solution:
Molarity of KOH = moles of KOH / volume of KOH solution (in L)
Given:
Volume of KOH solution = 80.0 mL = 80.0/1000 L = 0.0800 L
Molarity of KOH = 0.0651 moles / 0.0800 L = 0.81375 M
Therefore, the molarity of the KOH solution is approximately 0.81375 M.