Question

Saccharin is a weak organic base with a Kb of 4.80 x 10-3. A 0.297-g sample of saccharin dissolved in25.0 mL of water has a pH of 12.190. What is the molar mass of saccharin?A) 0.616 g/molB) 19.3 g/molC) 184 g/molD) 181 g/molE) 119 g/mol

Answer 1

Answer: The molar mass of the saccharin is 181 g/mol

Step-by-step explanation:

We are given:

pH of the solution = 12.190

To calculate pOH of the solution, we use the equation:

`pH+pOH=14\\pOH=14-12.190=1.81`

To calculate the
`OH^-` concentration, we use the equation:

`pOH=-\log[OH^-]\\\\1.81=-\log [OH^-]`

`[OH^-]=0.0155M`

As, saccharin is a weak base. Let us consider be as BOH

The equation for the reaction of BOH with water follows:

`BOH+H_2O\rightleftharpoons BH^++OH^-`

Initial: c

At eqllm: c-x x x

Value of x =
`[OH^-]=0.0155M`

The expression of
`K_b` for above equation follows:

`K_b=([BH^+]* [OH^-])/([BOH])`

We are given:

`K_b=4.80* 10^(-3)`

`[BH^+]=0.0155M`

`[OH^-]=0.0155M`

`[BOH]=c-0.0155`

Putting values in above expression, we get:

`4.80* 10^(-3)=(0.0155* 0.0155)/(c-0.0155)\\\\c=0.0655M`

Concentration of saccharin = 0.0655 M

To calculate the molecular mass of saccharin, 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 = 0.0655 M

Given mass of saccharin = 0.297 g

Volume of solution = 25.0 mL

Putting values in above equation, we get:

`0.0655M=\frac{0.297* 1000}{\text{Molar mass of saccharin}* 25.0}\\\\\text{Molar mass of saccharin}=(0.297* 1000)/(0.0655* 25)=181g/mol`

Hence, the molar mass of the saccharin is 181 g/mol