Question

Suppose that 14.56 ml of a 0.115 M solution NaOH (aq) was required to neutralize 0.0822g of an unknown triprotic acid, H3X. What is the molecular weight (g/mol) of the unknown acid?

What I have so far:

H3X + 3NaOH -> Na3X + 3H2O

0.0822/[0.115M (mol/L) NaOH × 0.01456L × (1molH3X/3molNaOH)] = 147g/mol H3X

What is the mass percent of protons of the unknown acid? <- This question please.

Thanks. ?

Answer 1

you calculated H3X molar mass = 147 g / mol

this means 147 g H3X has 3 x 1.00 g hydrogen

147 g H3X ---------------------> 3,0 g H

0.0822 g H3X ---------------> ?

mass of proton = 0.0822 x 3 / 147

                         = 1.67 x 10^-3 g

mass of protons = 1.67 x 10^-3 g

original sample mass = 0.0822 g

mass % of protons = protons mass x 100 / total sample mass

                             = 1.67 x 10^-3 x 100 / 0.0822

                         = 2.04 %