Question

0.5008 g of an unknown triprotic acid, H3A, is dissolved in 47.3 mL of water and then titrated with 0.315 M NaOH. It takes 25.72 mL of the NaOH solution to completely neutralize the acid. What is the molar mass of this acid

Answer 1

The molar mass is "185.44 g/mol".

Step-by-step explanation:

According to the question,

The moles of NaOH will be:

=
`(0.315)/(1000)* 25.72`

=
`8.1018* 10^(-3) \ moles`

Number of moles of an acid will be:

=
`(8.1018* 10^(-3))/(3)`

=
`2.7006* 10^(-3) \ moles`

We know that,

⇒
`Moles = (Mass)/(Molar \ mass)`

hence,

Molar mass of unknown acid will be:

=
`(Mass \ in \ g)/(Moles)`

=
`(0.5008)/(2.7006* 10^(-3))`

=
`185.44 \ g/mol`