How many ml of a 0.827 Molar KOH solution is required for neutralization in a titration of 35.00 ml of 0.737 Molar H2SO4?
Write the equation.
2KOH + H2SO4 ==> 2H2O + K2SO4
mols H2SO4 = M x L = 0.03500 L x 0.737 M = ??
mols KOH = 2 x mols H2SO4 from the equation.
mols KOH = M x L. You know mols KOH, you know M, solve for L and change to mL.
62.4 ml
thanks
You're welcome! I'm glad I could help. Just remember, chemistry may be titrating your brain cells, but with a little humor, we can neutralize the stress. Keep up the great work in your studies!
To find the volume of the 0.827 M KOH solution required for neutralization, we need to calculate the moles of H2SO4 and then use the stoichiometry of the balanced equation to determine the moles of KOH required.
Given:
- Volume of H2SO4 solution (V1) = 35.00 ml = 0.03500 L
- Molarity of H2SO4 solution (M1) = 0.737 M
- Molarity of KOH solution (M2) = 0.827 M
Step 1: Calculate the moles of H2SO4
mols H2SO4 = M1 x V1
mols H2SO4 = 0.737 M x 0.03500 L
mols H2SO4 = 0.0258 moles
Step 2: Use stoichiometry to determine the moles of KOH required
From the balanced equation: 2KOH + H2SO4 -> 2H2O + K2SO4
1 mole of H2SO4 reacts with 2 moles of KOH.
Therefore, mols KOH = 2 x mols H2SO4
mols KOH = 2 x 0.0258 moles
mols KOH = 0.0516 moles
Step 3: Calculate the volume of the KOH solution required
V2 = mols KOH / M2
V2 = 0.0516 moles / 0.827 M
V2 = 0.0624 L or 62.4 ml
Therefore, 62.4 ml of the 0.827 M KOH solution is required for neutralization in the titration.
To find the volume of the 0.827 Molar KOH solution required for neutralization, you first need to calculate the number of moles of H2SO4 present in the initial 35.00 ml of 0.737 Molar H2SO4 solution.
Using the formula Molarity (M) = Moles (mol) / Volume (L), you can calculate the number of moles of H2SO4:
Moles H2SO4 = M x L = 0.737 M x 0.03500 L = 0.025795 mol
From the balanced chemical equation, you can see that 2 moles of KOH react with 1 mole of H2SO4 to form 2 moles of H2O and 1 mole of K2SO4. This tells you that the number of moles of KOH required is twice the number of moles of H2SO4.
Moles KOH = 2 x Moles H2SO4 = 2 x 0.025795 mol = 0.05159 mol
Now, you can use the formula Moles (mol) = Molarity (M) x Volume (L) to find the volume of the 0.827 Molar KOH solution needed:
0.05159 mol = 0.827 M x Volume (L)
Solving for Volume (L), you get:
Volume (L) = 0.05159 mol / 0.827 M ≈ 0.0624 L
Finally, to convert the volume from liters to milliliters, multiply by 1000:
Volume (mL) = 0.0624 L x 1000 = 62.4 mL
Therefore, approximately 62.4 ml of the 0.827 Molar KOH solution is required for neutralization in the titration.