Benzene sulfonic acid, C6H5SO3H, is a strong acid and will neutralise sodium hydroxide solution (a 1:1 molar ratio). Benzene sulfonic acid crystals (y grams) were weighed into a titration flask and titrated against sodium hydroxide solution (1.0 M) and needed a volume of 26.2 cm3. What was the mass, y grams, of benzene sulfonic acid used (use C = 12, H = 1, S = 32, O = 16 amu)?

mol NaOH = M x L = ?

mols acid = mols NaOH
mols acid = grams/molar mass. You know molar mass and mols, solve for grams.

To find the mass of benzene sulfonic acid (C6H5SO3H), we need to use the given information of the volume used in the titration. Here are the steps to calculate the mass:

1. Determine the number of moles of sodium hydroxide (NaOH) used. Since the molar ratio between benzene sulfonic acid and sodium hydroxide is 1:1, the number of moles of NaOH is equal to the volume used in liters (26.2 cm³ = 26.2/1000 = 0.0262 L) multiplied by the molarity of the NaOH solution (1.0 M):

Moles of NaOH = Volume of NaOH (L) x Molarity of NaOH (mol/L)
= 0.0262 L x 1.0 mol/L
= 0.0262 mol NaOH

2. Since the molar ratio between benzene sulfonic acid and NaOH is 1:1, the number of moles of benzene sulfonic acid is also 0.0262 mol.

3. Calculate the molar mass of benzene sulfonic acid by adding the masses of each element in the molecule. According to the given amu values, C = 12 amu, H = 1 amu, S = 32 amu, and O = 16 amu:

Molar mass of C6H5SO3H = (6 x 12) + (5 x 1) + 32 + (3 x 16)
= 72 + 5 + 32 + 48
= 157 g/mol

4. Finally, calculate the mass of benzene sulfonic acid (y grams) by multiplying the number of moles of benzene sulfonic acid by its molar mass:

Mass of benzene sulfonic acid (y grams) = Moles of C6H5SO3H x Molar mass of C6H5SO3H
= 0.0262 mol x 157 g/mol
= 4.10 grams

Therefore, the mass of benzene sulfonic acid used in the titration is approximately 4.10 grams.