Final answer:
The molar mass of the unknown triprotic acid is calculated by dividing the mass of the acid (50.0 grams) by the number of moles of acid (0.331 moles), giving a molar mass of 151.06 g/mol.
Step-by-step explanation:
To determine the molar mass of the unknown triprotic acid, we need to first understand that a triprotic acid can donate three protons (H+ ions) per molecule of acid. The question tells us that 0.993 moles of NaOH were required to completely neutralize 50.0 grams of the acid. Since NaOH has a one-to-one molar ratio with each proton from the acid, we can write the neutralization reaction as:
H3A (aq) + 3NaOH (aq) → 3NaA (aq) + 3H2O (l)
To find the moles of the triprotic acid, we divide the moles of NaOH by 3, because it takes 3 moles of NaOH to neutralize 1 mole of triprotic acid:
Moles of triprotic acid = ÷ 0.993 moles NaOH ÷ 3 = 0.331 moles
Now we can find the molar mass of the unknown acid by dividing the mass by the moles:
Molar mass = ÷ 50.0 g / 0.331 moles = 151.06 g/mol
The molar mass of the unknown triprotic acid is 151.06 g/mol.