Question

50 cm³ of 0.2 n hcl is titrated against 0.1 n naoh solution. The titration was discontinued after adding 50 cm³ of naoh. The remaining titration is completed by adding 0.5 n koh. The volume of koh required for completing the titration is (karnataka, 2011)

Answer 1

Answer:

Explanation:

n1v1=n2v2

0.1*50=0.5*v2

0.1/0.5*50

50/5

10cm^3

Answer 2

The volume of KOH required for completing the titration is $10cm^3$

Explanation:

To calculate the volume of KOH, we use the equation given by neutralization reaction:

$N_1V_1=N_2V_2+N_3V_3$

where,

$N_1\text{ and }V_1$ are the normality and volume of acid which is HCl

$N_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

$N_3\text{ and }V_3$ are the n-factor, molarity and volume of base which is KOH

We are given:

$N_1=0.2N\\V_1=50cm^3\\N_2=0.1N\\V_2=50cm^3\\N_3=0.5N\\V_3=?cm^3$

Putting values in above equation, we get:

$0.2\times 50=(0.1\times 50)+(0.5\times V_3)\\\\V_3=\frac{(0.2\times 50)-(0.1\times 50)}{0.5}=10cm^3$

Learn more about neutralization reaction:

https://brainly.in/question/11899192

https://brainly.in/question/11716499

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