A volume of 60.0mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What was the molarity of the KOH solution if 19.2mL of 1.50 M H2SO4 was needed? The equation is
2KOH(aq)+H2SO4(aq)→K2SO4(aq)+2H2O(l)
Express the molar concentration numerically.
0.96mol/L
mols H2SO4 = M x L = ?
Using the coefficients in the balanced equation, convert mols H2SO4 to mols NaOH.
Then M NaOH = mols NaOH/L NaOH
To determine the molarity of the KOH solution, we can use the balanced chemical equation and the volume and concentration of the sulfuric acid.
From the balanced chemical equation: 2KOH(aq) + H2SO4(aq) → K2SO4(aq) + 2H2O(l), we can see that it takes 2 moles of KOH to react with 1 mole of H2SO4.
Given that 19.2 mL of 1.50 M H2SO4 was needed, we can calculate the number of moles of H2SO4 used as:
moles of H2SO4 = volume of H2SO4 (L) x concentration of H2SO4 (M)
moles of H2SO4 = 19.2 mL x 0.001 L/mL x 1.50 M
moles of H2SO4 = 0.0288 mol
Since the ratio of KOH to H2SO4 is 2:1, the number of moles of KOH used is also 0.0288 mol.
To calculate the molarity of the KOH solution, we use the equation:
molarity of KOH (M) = moles of KOH / volume of KOH (L)
molarity of KOH = 0.0288 mol / (60.0 mL x 0.001 L/mL)
molarity of KOH = 0.480 M
Therefore, the molarity of the KOH solution is 0.480 M.
To determine the molarity of the KOH solution, we can use the concept of stoichiometry and the balanced chemical equation provided.
First, let's identify the stoichiometry of the reaction. According to the balanced chemical equation:
2KOH(aq) + H2SO4(aq) → K2SO4(aq) + 2H2O(l)
The stoichiometric ratio between KOH and H2SO4 is 2:1. This means that for every 2 moles of KOH, we need 1 mole of H2SO4.
Since we know the volume and molarity of H2SO4 used, we can calculate the number of moles of H2SO4:
moles of H2SO4 = volume of H2SO4 (in L) x molarity of H2SO4
= 19.2 mL x (1 L / 1000 mL) x 1.50 M
= 0.0288 mol
Since the stoichiometric ratio is 2:1 between KOH and H2SO4, the number of moles of KOH is also 0.0288 mol.
Next, we can use the definition of molarity to calculate the molarity of the KOH solution:
Molarity = moles of solute / volume of solution (in L)
We are given that the volume of the KOH solution is 60.0 mL, which is equivalent to 0.0600 L.
Molarity of KOH = moles of KOH / volume of KOH solution
= 0.0288 mol / 0.0600 L
= 0.480 M
Therefore, the molarity of the KOH solution is 0.480 M (expressed numerically).