Question

If 10.62 mL of a standard 0.3330 M KOH solution reacts with 98.20 mL of CH3COOH solution, what is the molarity of the acid solution?

Answer 1

0.036 M of
`CH_(3) COOH`

Step-by-step explanation:

It is an example of acid-base neutralization reaction.

KOH +
`CH_(3) COOH` ---->
`CH_(3) COO^(-) K^(+)` +
`H_(2)O`

Base Acid Salt

When two component react then the number of moles of both the component should be same, therefore the number of moles and acids and bases should be the same in the following .

Molarity=
`\frac{\textrm{No. of Moles}}{\textrm{Volume of the Particular Solution}}`

No.of moles= Molarity × Volume of the Particular Solution

Therefore,

`M_(1)V_(1) =M_(2)V_(2)`------------------------------(1)

where

`M_(1)`= Molarity of Acid

`V_(1)`= Volume of Acid

`M_(2)`= Molarity of Base

`V_(2)`= Volume of Base

`M_(1)`=0.3330 M

`V_(1)`=10.62 mL

`V_(2)`=98.2 mL

`M_(2)`=??(in M)

Plugging in Equation 1,

0.3330 × 10.62 =
`M_(2)` × 98.2

`M_(2)`=
`(0.3330*10.62)/(98.2)`

`M_(2)`=0.036 M