Question

A monoprotic acid is an acid that donates a single proton to the solution. Suppose you have 0.140 g of a monoprotic acid dissolved in 35.0 mL of water. This solution is then neutralized with 14.5 mL of 0.110 M NaOH. What is the molar mass of the acid?

Answer 1

molar mass HA = 87.8 g/mole

Step-by-step explanation:

Given 0.140g HA + 14.5ml(0.110M NaOH) => NaA + H₂O

Rxn is a 1:1 rxn ration => moles HA neutralized = moles NaOH used

=> 0.140g/molar mass of HA = (0.110M)(0.0145L)

=> molar mass of HA = (0.140g)/[(0.110moles/L)(0.0145L)] = 87.8 g/mole

Answer 2

Answer: The molar mass of monoprotic acid is 87.72 g/mol

Step-by-step explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:

`n_1M_1V_1=n_2M_2V_2`

where,

`n_1,M_1\text{ and }V_1` are the n-factor, molarity and volume of monoprotic acid

`n_2,M_2\text{ and }V_2` are the n-factor, molarity and volume of base which is NaOH.

We are given:

`n_1=1\\M_1=?M\\V_1=35.0mL\\n_2=1\\M_2=0.110M\\V_2=14.5mL`

Putting values in above equation, we get:

`1* M_1* 35.0=1* 0.110* 14.5\\\\x=(1* 0.110* 14.5)/(1* 35.0)=0.0456M`

To calculate the molecular mass of acid, we use the equation used to calculate the molarity of solution:

`\text{Molarity of the solution}=\frac{\text{Mass of solute}* 1000}{\text{Molar mass of solute}* \text{Volume of solution (in mL)}}`

We are given:

Molarity of solution = 0.0456 M

Given mass of acid = 0.140 g

Volume of solution = 35.0 mL

Putting values in above equation, we get:

`0.0456M=\frac{0.140* 1000}{\text{Molar mass of acid}* 35}\\\\\text{Molar mass of acid}=(0.140* 1000)/(0.0456* 35)=87.72g/mol`

Hence, the molar mass of monoprotic acid is 87.72 g/mol