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?

21.33 g/mol

Explanation:

Considering:-

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,

Given  that:  

So,    

x = 0.93744 M

Also,

Considering:

Or,

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 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.

geosphere and hydrosphere.

this is because the geosphere has to do with geological events, while the hydrosphere controls all water on earth.