neutralize.py

Created by thabo
Created on January 06, 2025
1.59 KB
"""
Problem Statement

You need to determine how many
milliliters of 0.500 M sulfuric
acid (H2SO4) will be necessary
to neutralize 200 mL of 0.750 M
sodium hydroxide (NaOH) 
solution.

Steps to Solve the Problem

1. Write the Balanced Chemical
Equation

First, we write the chemical 
reaction that happens when 
sulfuric acid reacts with 
sodium hydroxide:
H2SO4 (aq) + 2NaOH (aq) 
-> Na2SO4 (aq) + 2H2O (l)

This equation shows that one
mole of sulfuric acid reacts
with two moles of sodium 
hydroxide to produce one mole 
of sodium sulfate and two moles
of water.

2. Calculate the Moles of 
Sodium Hydroxide (NaOH)

We use the formula for moles:
Moles = Molarity * Volume (L)

First, we need to convert the
volume from milliliters (mL) 
to liters (L):
200 mL = 0.200 L

Now, calculate the moles of 
NaOH:
Moles of NaOH = 
  0.750 M * 0.200 L = 
  0.150 moles

3. Determine the Moles of 
Sulfuric Acid (H2SO4) Needed

From the balanced equation, 
we see that 1 mole of H2SO4 
reacts with 2 moles of NaOH. 
This means we need half as 
many moles of H2SO4 as we have
moles of NaOH.

Moles of H2SO4 = Moles of NaOH / 2
Moles of H2SO4 = 
  0.150 moles / 2 = 0.075 moles

4. Calculate the Volume of 
Sulfuric Acid (H2SO4) Needed

We need to find out how many 
milliliters of 0.500 M H2SO4 
solution contains 0.075 moles.
Use the formula for volume:
Volume (L) = Moles / Molarity

Volume of H2SO4 (L) = 
  0.075 moles / 0.500 M = 0.150 L

Convert the volume from liters
to milliliters:
0.150 L = 150 mL

Summary

To neutralize 200 mL of 0.750 M
NaOH solution, you will need
150 mL of 0.500 M H2SO4 
solution.

"""