Automobile batteries use 3.0 M H2SO4 as an electrolyte. How much 1.20 M NaOH will be needed to neutralize 225mL of battery acid?
1.1
To find out how much 1.20 M NaOH is needed to neutralize 225 mL of battery acid, we first need to determine the number of moles of H2SO4 present in the battery acid.
Step 1: Convert the volume of battery acid to liters.
225 mL = 225/1000 = 0.225 L
Step 2: Calculate the number of moles of H2SO4 using the molarity and volume of the battery acid.
Moles of H2SO4 = Molarity x Volume
Moles of H2SO4 = 3.0 M x 0.225 L
Step 3: Determine the number of moles of NaOH required to neutralize the moles of H2SO4.
The balanced equation for the neutralization reaction between H2SO4 and NaOH is 1:2. This means that one mole of H2SO4 reacts with two moles of NaOH.
Moles of NaOH required = 2 x Moles of H2SO4
Step 4: Calculate the volume of 1.20 M NaOH required using its molarity and the number of moles of NaOH.
Moles of NaOH = Moles of NaOH required
Volume of NaOH = Moles of NaOH / Molarity
Let's calculate the solution now:
Moles of H2SO4 = 3.0 M x 0.225 L = 0.675 mol H2SO4
Moles of NaOH required = 2 x 0.675 mol H2SO4 = 1.35 mol NaOH
Volume of NaOH = 1.35 mol / 1.20 M = 1.125 L = 1125 mL
Therefore, 1.20 M NaOH will be needed to neutralize 1125 mL of battery acid.
To find out how much 1.20 M NaOH is needed to neutralize 225 mL of battery acid, we need to determine the number of moles of H2SO4 present in the 225 mL of battery acid.
First, we convert the volume from mL to L:
225 mL = 225/1000 = 0.225 L
Next, we need to determine the number of moles of H2SO4. Using the given concentration of 3.0 M, we can calculate it using the formula:
moles = concentration (M) × volume (L)
moles of H2SO4 = 3.0 M × 0.225 L = 0.675 moles
Since the reaction between H2SO4 and NaOH is a 1:2 ratio, it means that we will need twice the number of moles of NaOH to neutralize the H2SO4. Therefore, we need:
moles of NaOH = 2 × moles of H2SO4 = 2 × 0.675 = 1.35 moles
Finally, we need to find the volume of 1.20 M NaOH solution required to contain 1.35 moles of NaOH. We can calculate this using the formula:
volume (L) = moles / concentration (M)
volume of 1.20 M NaOH = 1.35 moles / 1.20 M ≈ 1.125 L
Therefore, approximately 1.125 L or 1125 mL of 1.20 M NaOH will be needed to neutralize 225 mL of battery acid.
2NaOH + H2SO4 ==> Na2SO4 + 2H2O
mols H2SO4 = M x L = ?
Using the coefficients in the balanced e4quatioin, convert mols H2SO4 to mols NaOH.
Now M NaOH = mols NaOH/L NaOH. You know mols and M, solve for L NaOH and conert to mL if you wish.