Question

A 0.623 g sample of a monoprotic acid is dissolved in water and titrated with 0.260 M KOH.

What is the molar mass of the acid if 19.0 mL of the KOH solution is required to neutralize the sample?

Answer 1

126. g/mol is the molar mass of the acid .

Step-by-step explanation:

`HA+KOH\rightarrow KA+H_2O`

Moles of KOH = n

Volume of the solution of KOH = 19.0 mL = 0.019 L

Molarity of the KOH solution = 0.260 M

`Molarity=(Moles)/(Volume(L))`

`0.260M=(n)/(0.019 L)`

`n=0.260 M* 0.019 L=0.00494 mol`

Accrding to recation, 1 mol KOH react with 1 mole of HA ,then 0.00494 mol of KOH will react with :

`(1)/(1)* 0.00494 mol=0.00494 mol` of HA

Mass of HA = 0.623 g

Molar mass of HA = M

`0.00494 mol=(0.623 g)/(M)`

[tex[M=\frac{0.623 g}{0.00494 mol}=126. g/mol[/tex]

126. g/mol is the molar mass of the acid .