100 g of water is added into 10.6 m NaOH solution. The diluted NaOH solution is then neutralized completely with H2SO4 solution. Density of NaOH is 1.08 g/ml. Calculate:
i. the mol fraction of NaOH in the diluted NaOH solution.
ii. mol of H2SO4 required.
how many moles of NaOH in 1.06 g?
half that many moles of H2SO$, since
H2SO4 + 2NaOH = Na2SO4 + 2H2O
To calculate the mol fraction of NaOH in the diluted NaOH solution, we need to determine the moles of NaOH and the total moles of solute and solvent in the solution.
Step 1: Calculate the moles of NaOH:
Given that the density of NaOH is 1.08 g/mL, we can calculate the mass of NaOH in the solution:
mass of NaOH = volume of NaOH solution x density of NaOH
= 10.6 mL x 1.08 g/mL
= 11.448 g
Now, we can determine the moles of NaOH using its molar mass:
moles of NaOH = mass of NaOH / molar mass of NaOH
= 11.448 g / 40.00 g/mol (molar mass of NaOH)
= 0.2862 mol
Step 2: Calculate the total moles of solute and solvent:
We have 100 g of water added to the solution. Water has a molar mass of 18.015 g/mol.
moles of water = mass of water / molar mass of water
= 100 g / 18.015 g/mol
= 5.548 mol
Total moles of solute and solvent = moles of NaOH + moles of water
= 0.2862 mol + 5.548 mol
= 5.8342 mol
Step 3: Calculate the mol fraction of NaOH:
mol fraction of NaOH = moles of NaOH / total moles of solute and solvent
= 0.2862 mol / 5.8342 mol
= 0.0491
Therefore, the mol fraction of NaOH in the diluted NaOH solution is approximately 0.0491.
Now, let's calculate the moles of H2SO4 required to neutralize the diluted NaOH solution.
Using the balanced equation for the neutralization reaction between NaOH and H2SO4:
2NaOH + H2SO4 -> Na2SO4 + 2H2O
The equation shows that it requires 2 moles of NaOH to neutralize 1 mole of H2SO4.
Step 4: Calculate the moles of H2SO4 required:
moles of H2SO4 = 2 x moles of NaOH
= 2 x 0.2862 mol
= 0.5724 mol
Therefore, approximately 0.5724 mol of H2SO4 is required to completely neutralize the diluted NaOH solution.