What volume of 0.5 N nitric acid would be required to completely neutralize 300 ml of 1.0 N sodium hydroxide?
1 mol of HNO3 will neutralise 1 mol of NaOH
since concentration of HNO3 is half that of NaOH, volume will have to double. So 600ml
To find the volume of 0.5 N nitric acid required to neutralize 300 ml of 1.0 N sodium hydroxide, we need to use the concept of acid-base neutralization and the equation of mole-to-mole ratio.
Here is how to calculate it step by step:
Step 1: Determine the balanced chemical equation for the reaction between nitric acid (HNO3) and sodium hydroxide (NaOH):
HNO3 + NaOH → NaNO3 + H2O
This equation shows that one mole of nitric acid reacts with one mole of sodium hydroxide to produce one mole of sodium nitrate and one mole of water.
Step 2: Calculate the number of moles of sodium hydroxide:
Molarity (M) = Moles (n) / Volume (V)
For the given solution of 1.0 N sodium hydroxide,
Molarity (NaOH) = 1.0 N = 1.0 mol/L
Volume of sodium hydroxide (V) = 300 ml = 300/1000 = 0.3 L
So, Moles of sodium hydroxide (n) = Molarity (M) x Volume (V)
n(NaOH) = 1.0 mol/L x 0.3 L = 0.3 moles of NaOH
Step 3: Find the mole-to-mole ratio between nitric acid and sodium hydroxide from the balanced equation:
From the balanced equation, we can see that 1 mole of nitric acid reacts with 1 mole of sodium hydroxide.
Step 4: Determine the volume of 0.5 N nitric acid:
Moles of nitric acid = Moles of sodium hydroxide (n(NaOH))
Let's assume the volume of 0.5 N nitric acid required is V.
Molarity (HNO3) = 0.5 N = 0.5 mol/L
Moles of nitric acid (n(HNO3)) = Molarity (M) x Volume (V)
n(HNO3) = 0.5 mol/L x V
Since the mole-to-mole ratio between nitric acid and sodium hydroxide is 1:1, we have:
n(HNO3) = 0.5 mol/L x V = 0.3 moles of NaOH
Now we can set up an equation to solve for V:
0.5 V = 0.3
Dividing both sides of the equation by 0.5:
V = 0.3 / 0.5
V = 0.6 L or 600 ml
Therefore, 600 ml of 0.5 N nitric acid would be required to completely neutralize 300 ml of 1.0 N sodium hydroxide.