A storage tank containing 1.6x10^4 Liters of water is found to be contaminated with nitric acid and therefore has a pH of 3.78. Calculate the volume of 0.050 M sodium hydroxide required to be added to the tank in order to neutralize (pH = 7.0) the contaminated water.
HNO3 + NaOH ==> NaNO3 + HOH
Notice ratio acid/base = 1:1
pH 3.78 = (H^+) = (HNO3) = 1.7E-4 (approximate--you should do it more accurately.)
1.7E-4 moles/L x 1.6E4L = about 3 moles = moles HNO3 = moles NaOH needed.
M NaOH = moles NaOH/L NaOH. You have M and moles, solve for L.
To calculate the volume of sodium hydroxide required to neutralize the contaminated water, we need to first determine the number of moles of acid present in the tank.
Step 1: Convert the volume of water to liters
The volume of water is given as 1.6x10^4 Liters. Since the standard unit for volume in calculations is liters, no conversion is required in this step.
Step 2: Calculate the number of moles of acid
To calculate the number of moles of acid, we need to use the equation for pH:
pH = -log[H+]
where [H+] represents the concentration of hydrogen ions in moles per liter.
In this case, we are given the pH as 3.78. To find [H+], we use the equation:
[H+] = 10^(-pH)
[H+] = 10^(-3.78)
[H+] = 1.58x10^(-4) moles per liter
Now, to calculate the number of moles in the entire tank, we need to multiply the concentration by the volume:
moles of acid = [H+] x volume of water
moles of acid = (1.58x10^(-4) moles/L) x (1.6x10^4 L)
moles of acid = 2.53 moles
Step 3: Convert moles of acid to moles of sodium hydroxide
Since nitric acid (HNO3) is a strong acid and sodium hydroxide (NaOH) is a strong base, the reaction between them is 1:1 in terms of moles. Therefore, the number of moles of sodium hydroxide required to neutralize the acid is also 2.53 moles.
Step 4: Calculate the volume of sodium hydroxide
To calculate the volume of sodium hydroxide, we need to use the equation:
volume (L) = moles / concentration (M)
In this case, the concentration given is 0.050 M (0.050 moles/L).
volume of sodium hydroxide = 2.53 moles / 0.050 moles/L
volume of sodium hydroxide = 50.6 Liters
Therefore, 50.6 Liters of 0.050 M sodium hydroxide are required to neutralize the contaminated water in the tank.