Suppose that 16.07 mL of 0.0512 M NaOH were required to titrate a sample of unknown acid. How many moles of NaOH were used? Assuming that the unknown acid sample in question 1 had a mass of 0.177 g, what is the molar mass of the unknown acid?
Is this a monoprotic acid or otherwise?
I assume monoprotic.
mols NaOH = M x L = ?
mols acid = mols NaOH (if acid is monoprotic)
mols acid = grams/molar mass.
You know grams of the acid and mols of the acid, solve for molar mass.
To find the number of moles of NaOH used, we can use the equation:
moles of solute (NaOH) = concentration (M) x volume (L)
Given that the concentration of NaOH is 0.0512 M and the volume used is 16.07 mL (which can be converted to L by dividing by 1000), we can calculate:
moles of NaOH = 0.0512 M x (16.07 mL / 1000 mL/L)
moles of NaOH = 0.0512 M x 0.01607 L
moles of NaOH ≈ 0.000821 moles
So, approximately 0.000821 moles of NaOH were used.
To find the molar mass of the unknown acid, we can use the equation:
molar mass = mass (g) / moles
Given that the mass of the unknown acid is 0.177 g and we have already determined the moles of NaOH used, we can substitute these values into the equation:
molar mass = 0.177 g / 0.000821 moles
molar mass ≈ 215.26 g/mol
Therefore, the molar mass of the unknown acid is approximately 215.26 g/mol.