<u>Answer:</u> The volume of NaOH required is 402.9 mL
<u>Explanation:</u>
To calculate the number of moles for given molarity, we use the equation:
.....(1)
Molarity of HCl solution = 0.315 M
Volume of solution = 503.4 mL = 0.5034 L (Conversion factor: 1 L = 1000 mL)
Putting values in equation 1, we get:
![0.315M=\frac{\text{Moles of HCl}}{0.5034L}\\\\\text{Moles of HCl}=(0.315mol/L\times 0.5034L)=0.1586mol](https://tex.z-dn.net/?f=0.315M%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20HCl%7D%7D%7B0.5034L%7D%5C%5C%5C%5C%5Ctext%7BMoles%20of%20HCl%7D%3D%280.315mol%2FL%5Ctimes%200.5034L%29%3D0.1586mol)
- <u>For sulfuric acid:</u>
Molarity of sulfuric acid solution = 0.125 M
Volume of solution = 503.4 mL = 0.5034 L
Putting values in equation 1, we get:
![0.125M=\frac{\text{Moles of }H_2SO_4}{0.5034L}\\\\\text{Moles of }H_2SO_4=(0.125mol/L\times 0.5034L)=0.0630mol](https://tex.z-dn.net/?f=0.125M%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20%7DH_2SO_4%7D%7B0.5034L%7D%5C%5C%5C%5C%5Ctext%7BMoles%20of%20%7DH_2SO_4%3D%280.125mol%2FL%5Ctimes%200.5034L%29%3D0.0630mol)
As, all of the acid is neutralized, so moles of NaOH = [0.1586 + 0.0630] moles = 0.2216 moles
Molarity of NaOH solution = 0.55 M
Moles of NaOH = 0.2216 moles
Putting values in equation 1, we get:
![0.55M=\frac{0.2216}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{0.2216}{0.55}=0.4029L=402.9mL](https://tex.z-dn.net/?f=0.55M%3D%5Cfrac%7B0.2216%7D%7B%5Ctext%7BVolume%20of%20solution%7D%7D%5C%5C%5C%5C%5Ctext%7BVolume%20of%20solution%7D%3D%5Cfrac%7B0.2216%7D%7B0.55%7D%3D0.4029L%3D402.9mL)
Hence, the volume of NaOH required is 402.9 mL