Home
Class 12
CHEMISTRY
100 ml of 1M NaOH is mixed with 50ml of ...

100 ml of 1M NaOH is mixed with 50ml of 1N KOH solution. Normality of mixture is

A

1N

B

0.5 N

C

0.25 N

D

2 N

Text Solution

AI Generated Solution

The correct Answer is:
To find the normality of the mixture when 100 ml of 1M NaOH is mixed with 50 ml of 1N KOH, we can follow these steps: ### Step 1: Calculate the millimoles of NaOH - **Formula**: Millimoles = Molarity (M) × Volume (ml) - For NaOH: - Molarity = 1 M - Volume = 100 ml - Calculation: \[ \text{Millimoles of NaOH} = 1 \, \text{M} \times 100 \, \text{ml} = 100 \, \text{mmol} \] ### Step 2: Calculate the millimoles of KOH - **Formula**: Millimoles = Normality (N) × Volume (ml) - For KOH: - Normality = 1 N - Volume = 50 ml - Calculation: \[ \text{Millimoles of KOH} = 1 \, \text{N} \times 50 \, \text{ml} = 50 \, \text{mmol} \] ### Step 3: Calculate the total millimoles in the mixture - Total millimoles = Millimoles of NaOH + Millimoles of KOH - Calculation: \[ \text{Total millimoles} = 100 \, \text{mmol} + 50 \, \text{mmol} = 150 \, \text{mmol} \] ### Step 4: Calculate the total volume of the mixture - Total volume = Volume of NaOH + Volume of KOH - Calculation: \[ \text{Total volume} = 100 \, \text{ml} + 50 \, \text{ml} = 150 \, \text{ml} \] ### Step 5: Calculate the molarity of the mixture - **Formula**: Molarity (M) = Total millimoles / Total volume (in liters) - Convert volume from ml to liters: \[ 150 \, \text{ml} = 0.150 \, \text{L} \] - Calculation: \[ \text{Molarity} = \frac{150 \, \text{mmol}}{150 \, \text{ml}} = \frac{150 \, \text{mmol}}{0.150 \, \text{L}} = 1 \, \text{M} \] ### Step 6: Determine the normality of the mixture - Since both NaOH and KOH have a valency of 1 (n=1), the normality is equal to the molarity. - Therefore, the normality of the mixture is: \[ \text{Normality} = 1 \, \text{N} \] ### Final Answer: The normality of the mixture is **1 N**. ---
Promotional Banner

Similar Questions

Explore conceptually related problems

100 ml of 0.1 N NaOH is mixed with 50 ml of 0.1 N H_2SO_4 . The pH of the resulting solution is

If 100 mL of 1 N H_(2)SO_(4) is mixed with 100 mL of 1 M NaOH solution. The resulting solution will be

"40 ml "(N)/(10)HCl solution is mixed with 60 ml of (N)/(20)KOH solution. The resulting mixture will be

20 mL of 0.1N HCI is mixed with 20 ml of 0.1N KOH . The pH of the solution would be

50ml N//10 NaOH solution is mixed with 50 ml N//20 HCl solution. The resulting solution will (a) Turns phenolphthalein solution pink " " (b) Turns blue litmus red (c ) Turns methyl orange red " "(d) [H^(+)] & [OH^(-)]

When 100 ml of M/10 NaOH solution and 50 ml of M/5 HCI solution are mixed, the pH of resulting solution would be

A 50 ml solution of pH=1 is mixed with a 50 ml solution of pH=2 . The pH of the mixture will be nearly

When 5.0 mL of a 1.0 M HCl solution is mixed with 5.0 mL of a 0.1 M NaOH solution, temperature of solution is increased by 2^@C predicted accurately from this observation?

100mL, 3% (w/v) NaOH solution is mixed with 100 ml, 9% (w/v) NaOH solution. The molarity of final solution is

100ml 0.1M H_(3)PO_(4) is mixed with 50ml of 0.1M NaOH. What is the pH of the resultant solution? (Successive dissociation constant of H_(3) PO_(4) "are" 10^(-3) , 10^(-8) and 10^(-12) respectively ?