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If [tex]0.4 \, N \, 500 \, \text{ml} \, \text{HCl}[/tex] neutralizes [tex]300 \, \text{ml}[/tex] of [tex]\text{KOH}[/tex] completely, find the normality of the [tex]\text{KOH}[/tex] solution.

Sagot :

GinnyAnswer
To find the normality of the KOH solution, we can use the concept of chemical equivalence and the formula:

[tex]\[ N_1 \times V_1 = N_2 \times V_2 \][/tex]

Here:
- \( N_1 \) and \( V_1 \) are the normality and volume of the neutralizing solution (neutralizer),
- \( N_2 \) and \( V_2 \) are the normality and volume of the KOH solution.

Let's write down the given values:
- The normality of the neutralizer ( \( N_1 \) ) = 0.4 N
- The volume of the neutralizer ( \( V_1 \) ) = 500 ml
- The volume of the KOH solution ( \( V_2 \) ) = 300 ml

First, let's convert the volumes from milliliters to liters:
- The volume of the neutralizer in liters ( \( V_1 \) ) = \( \frac{500}{1000} \) = 0.5 L
- The volume of the KOH solution in liters ( \( V_2 \) ) = \( \frac{300}{1000} \) = 0.3 L

Now we can substitute these values into the formula to find the normality of the KOH solution ( \( N_2 \) ):
[tex]\[ N_1 \times V_1 = N_2 \times V_2 \][/tex]
[tex]\[ 0.4 \times 0.5 = N_2 \times 0.3 \][/tex]

Solving for \( N_2 \):
[tex]\[ 0.2 = N_2 \times 0.3 \][/tex]
[tex]\[ N_2 = \frac{0.2}{0.3} \][/tex]
[tex]\[ N_2 = \frac{2}{3} \][/tex]
[tex]\[ N_2 = 0.6667 \text{ N} \][/tex]

Therefore, the normality of the KOH solution is [tex]\( 0.6667 \text{ N} \)[/tex].
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