Answer: The molarity of the NaOH solution is 0.5 M.
Explanation:
Moles of
in 0.25 L of 2.00 M solution:
![Molarity=\frac{\text{Moles of }H_2SO_4}{\text{Volume in Liters of}H_2SO_4}](https://tex.z-dn.net/?f=Molarity%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20%7DH_2SO_4%7D%7B%5Ctext%7BVolume%20in%20Liters%20of%7DH_2SO_4%7D)
![2.00 M=\frac{\text{Moles of }H_2SO_4}{0.25 L}](https://tex.z-dn.net/?f=2.00%20M%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20%7DH_2SO_4%7D%7B0.25%20L%7D)
Moles of
= 2.00 mol/L × 0.25 L = 0.5 moles
Moles of
in 2.00 L of an unknown Molarity :
![H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O](https://tex.z-dn.net/?f=H_2SO_4%2B2NaOH%5Crightarrow%20Na_2SO_4%2B2H_2O)
1 mole
gives 2 moles of
ions.
Then 0.5 mole
will give: 0.5 mol × 2 = 1.0 mol
1.0 mol of
will neutralize the 1.0 mol of
ions.
Moles of
= Moles of
= 1.0 moles
1 mol of
are produced by 1 mol of NaOH
The 1 mole of
will be produced from = 1 mol of NaOH
![Molarity=\frac{\text{Moles of }NaOH}{\text{Volume in Liters of NaOH}}](https://tex.z-dn.net/?f=Molarity%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20%7DNaOH%7D%7B%5Ctext%7BVolume%20in%20Liters%20of%20NaOH%7D%7D)
![M=\frac{1 mol}{2.00L}=0.5 mol/L=0.5M](https://tex.z-dn.net/?f=M%3D%5Cfrac%7B1%20mol%7D%7B2.00L%7D%3D0.5%20mol%2FL%3D0.5M)
The molarity of the NaOH solution is 0.5 M.