What volume of 1.00 M NaOH would be required to neutralize 77.4 mL of 5.4 M HCl? Round to two places after the decimal point and keep it in mL
To calculate the volume of 1.00 M NaOH required to neutralize 77.4 mL of 5.4 M HCl, we can use the equation:
M1V1 = M2V2
Where:
M1 is the molarity of the NaOH solution
V1 is the volume of the NaOH solution
M2 is the molarity of the HCl solution
V2 is the volume of the HCl solution
In this case:
M1 = 1.00 M (Molarity of NaOH)
V1 = Volume of NaOH solution (unknown)
M2 = 5.4 M (Molarity of HCl)
V2 = 77.4 mL (Volume of HCl solution)
Rearranging the formula, we can solve for V1:
V1 = (M2 * V2) / M1
V1 = (5.4 M * 77.4 mL) / 1.00 M
V1 = 415.56 mL
Therefore, the volume of 1.00 M NaOH required to neutralize 77.4 mL of 5.4 M HCl is 415.56 mL. Rounded to two decimal places, the answer is 415.56 mL.
To solve this problem, we need to use the concept of stoichiometry and the equation for the neutralization reaction between HCl and NaOH. The balanced chemical equation is:
HCl + NaOH → NaCl + H2O
From the balanced equation, we can see that 1 mole of HCl reacts with 1 mole of NaOH. Therefore, the ratio of their volumes is also 1:1.
Here are the steps to find the volume of 1.00 M NaOH required to neutralize 77.4 mL of 5.4 M HCl:
Step 1: Determine the moles of HCl
Use the given volume and molarity of HCl to calculate the moles of HCl using the formula:
moles = volume (L) × molarity
First, convert the volume of HCl from milliliters (mL) to liters (L):
77.4 mL ÷ 1000 mL/L = 0.0774 L
Now, calculate the moles of HCl:
moles of HCl = 0.0774 L × 5.4 mol/L = 0.41796 mol (rounded to 5 decimal places)
Step 2: Calculate the volume of NaOH
Since the moles of HCl and NaOH are equal, we can use the molarity of NaOH to calculate the volume of NaOH needed using the formula:
volume (L) = moles / molarity
volume of NaOH = 0.41796 mol / 1.00 mol/L = 0.41796 L
Finally, convert the volume from liters back to milliliters:
volume of NaOH = 0.41796 L × 1000 mL/L = 417.96 mL
Round the answer to two decimal places, keeping it in mL:
Volume of 1.00 M NaOH required = 417.96 mL (rounded to two decimal places)
Therefore, 417.96 mL of 1.00 M NaOH would be required to neutralize 77.4 mL of 5.4 M HCl.
mols HCl = M x L = ?
mols NaOH = mols HCl
Then M NaOH = mols NaOH/L NaOH