What volume of 0.5M HNO3 is required to neutralize 25 cm3 of a 0.2M NaOH solution?
please show work and detail.
See above answer by Bob Pursley.
jgugiub
To find the volume of 0.5M HNO3 required to neutralize 25 cm3 of a 0.2M NaOH solution, we can use the concept of stoichiometry.
Step 1: Write and balance the chemical equation for the reaction:
NaOH + HNO3 → NaNO3 + H2O
This balanced equation tells us that 1 mole of NaOH reacts with 1 mole of HNO3.
Step 2: Calculate the number of moles of NaOH in 25 cm3 of the 0.2M solution.
Moles = Volume (in liters) × Concentration (in mol/L)
Moles of NaOH = (25 cm3 / 1000 cm3/L) × 0.2 mol/L
Moles of NaOH = 0.005 moles
Step 3: According to the balanced equation, the stoichiometric ratio is 1:1 between NaOH and HNO3. Therefore, the number of moles of HNO3 required is also 0.005 moles.
Step 4: Calculate the volume of 0.5M HNO3 required.
Volume (in liters) = Moles / Concentration (in mol/L)
Volume of HNO3 = 0.005 moles / 0.5 mol/L
Volume of HNO3 = 0.01 liters
Step 5: Convert the volume to the desired unit.
1 liter = 1000 cm3
Volume of HNO3 = 0.01 liters × 1000 cm3/L
Volume of HNO3 = 10 cm3
Therefore, 10 cm3 of 0.5M HNO3 is required to neutralize 25 cm3 of a 0.2M NaOH solution.