Question

You are given a 1.00 g sample of an unknown tri-protic acid, which you dissolve in 50.0 mL of water containing phenolphthalein indicator. You titrate the acid solution with standardized 0.400 M KOH(aq). It requires 39.06 mL of the KOH solution to produce a light pink indicator color. What is the molecular weight of the unknown acid?

Answer 1

21.33 g/mol

Step-by-step explanation:

Considering:-

`Normality=Molarity* {n-factor}`

So, Given that:-

Let Molarity of tri-protic acid = x M

n-factor of tri-protic acid = 3 (3 dissociable H)

So,

Normality of tri-protic acid = x/3 N

So, Given that:-

Let Molarity of KOH = 0.400 M

n-factor of KOH = 1 (1 dissociable OH)

So,

Normality of KOH = 0.400 N

Considering:-

At equivalence point

Gram equivalents of acid = Gram equivalents of base

So,

`Normality_(acid)* Volume_(acid)=Normality_(base)* Volume_(base)`

Given that:

`Normality_(base)=0.400\ N`

`Volume_(base)=39.06\ mL`

`Volume_(acid)=50.0\ mL`

`Normality_(acid)=x/3\ N`

So,

`(x)/(3)* 50=0.400* 39.06`

x = 0.93744 M

Also,

Considering:

`Molarity=(Moles\ of\ solute)/(Volume\ of\ the\ solution)`

Or,

`Moles =Molarity * {Volume\ of\ the\ solution}`

Given :

For tri-protic acid :

Molarity = 0.93744 M

Volume = 50.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 50.0×10⁻³ L

Thus, moles of tri-protic acid :

`Moles=0.93744 * {50.0* 10^(-3)}\ moles`

Moles of tri-protic acid = 0.046872 moles

Also, Given mass = 1.00 g

So,

Molar mass = Mass/Moles = 1.00g / 0.046872 moles = 21.33 g/mol

21.33 g/mol is the molecular weight of the unknown acid.