How much 6M HNO3 will be required to decrease the pH of 5000L solution of pH=9.6 to pH=3?
To calculate the amount of 6M HNO3 needed to decrease the pH of a solution, we can follow these steps:
Step 1: Find the change in pH:
The change in pH is the difference between the initial pH and the final pH. In this case, the change in pH is:
ΔpH = Final pH - Initial pH = 3 - 9.6 = -6.6
Step 2: Convert the volume of the solution to liters:
The given volume is 5000 L, which is already in liters.
Step 3: Calculate the moles of H+ ions required to achieve the desired change in pH:
The relationship between pH and the concentration of H+ ions in a solution is given by the equation:
pH = -log[H+]
[H+] = 10^(-pH)
Since the concentration of HNO3 is 6M, which means it is a 6 moles per liter solution, we can use this equation to calculate the moles of H+ ions required to achieve the desired change in pH:
Moles of H+ ions required = 10^(-ΔpH) = 10^(-(-6.6)) = 2.5119 × 10^6 moles
Step 4: Calculate the volume of 6M HNO3 needed:
To calculate the volume of 6M HNO3 needed, we can use the equation:
Volume of 6M HNO3 = Moles of H+ ions required / Concentration of HNO3
Volume of 6M HNO3 = (2.5119 × 10^6 moles) / (6 moles/L) = 418,650 L
Therefore, approximately 418,650 liters of 6M HNO3 will be required to decrease the pH of 5000 L solution from pH 9.6 to pH 3.