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20.0 ml sample of HCl, ist te of Hel is titrated with 0.0300M KOH solution. The volume of potassium droxide required is 27.50mL. What is the molarity of HCI solution? +KOH → KCI+H2O

User Joerage
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1 Answer

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Answer: The molarity of HCl is 0.041 M.

Step-by-step explanation:

To calculate the molarity 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 acid which is HCl


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

We are given:


n_1=1\\M_1=?M\\V_1=20.0mL\\n_2=1\\M_2=0.0300M\\V_2=27.50mL

Putting values in above equation, we get:


1* M_1* 20.0=1* 0.0300* 27.50\\\\M_1=0.041M

Hence, the molarity of HCl is 0.041 M.

User Thibauts
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