Answer:
molar mass HA = 87.8 g/mole
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: The molar mass of monoprotic acid is 87.72 g/mol
Explanation:
To calculate the concentration of acid, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of monoprotic acid
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is NaOH.
We are given:
[tex]n_1=1\\M_1=?M\\V_1=35.0mL\\n_2=1\\M_2=0.110M\\V_2=14.5mL[/tex]
Putting values in above equation, we get:
[tex]1\times M_1\times 35.0=1\times 0.110\times 14.5\\\\x=\frac{1\times 0.110\times 14.5}{1\times 35.0}=0.0456M[/tex]
To calculate the molecular mass of acid, we use the equation used to calculate the molarity of solution:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]
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:
[tex]0.0456M=\frac{0.140\times 1000}{\text{Molar mass of acid}\times 35}\\\\\text{Molar mass of acid}=\frac{0.140\times 1000}{0.0456\times 35}=87.72g/mol[/tex]
Hence, the molar mass of monoprotic acid is 87.72 g/mol
