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zepelin [54]
3 years ago
11

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 solut

ion?
Chemistry
1 answer:
Nikolay [14]3 years ago
3 0

Answer:

0.036 M of CH_{3} COOH

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}=\frac{0.3330*10.62}{98.2}

M_{2}=0.036 M

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