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.
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]
