How many grams of nitric acid (HNO3) must be dissolved in 590 ml of water to make a solution with a pOH = 10.1?
Select one:
a. 2.9 x 10-9 g
b. 7.9 x 10-3 g
c. 4.7 x 10-3 g
d. 7.9 x 10-6 g
e. 7.4 x 10-5 g
Convert pOH to pH with
pH + pOH = pKw = 14.
Convert pH to (H^+) with pH = -log(H^+).
Then since HNO3 is 100% ionized, H^+) = (HNO3)
mols HNO3 needed = M x L = ?
mols HNO3 = grams/molar mass. You know mols and L, solve for grams.
To determine the number of grams of nitric acid (HNO3) required, we first need to find the concentration of hydronium ions (H3O+) in the solution using the pOH value.
The formula to convert pOH to concentration of hydroxide ions is:
pOH = -log[OH-]
Since pOH is given as 10.1, we can calculate the concentration of hydroxide ions (OH-) using the equation:
10.1 = -log[OH-]
To find OH-, we need to take the inverse log of both sides:
[OH-] = 10^-(pOH)
Substituting the given pOH value:
[OH-] = 10^-(10.1)
Now, we can use the fact that for a strong acid like HNO3, the concentration of hydronium ions is equal to the concentration of nitric acid.
Therefore, the concentration of HNO3 = [H3O+] = [OH-] = 10^-(10.1)
To find the number of moles of HNO3, we need to convert its concentration to molarity using the volume of the solution.
Molarity (M) = moles of solute / liters of solution
The volume of the solution is given as 590 ml, which is equivalent to 0.59 liters.
Therefore, moles of HNO3 = Molarity x Volume
moles of HNO3 = [H3O+] x 0.59
Now, we can calculate the grams of HNO3 using its molar mass.
The molar mass of HNO3:
H = 1.01 g/mol
N = 14.01 g/mol
O = 16.00 g/mol (x3 since there are 3 O atoms)
Molar mass of HNO3 = 1.01 + 14.01 + (16.00 x 3) = 63.01 g/mol
Finally, we can calculate the grams of HNO3 using the moles of HNO3:
grams of HNO3 = moles of HNO3 x molar mass of HNO3
Step-by-step-wise, the calculations are as follows:
1. Calculate the concentration of hydroxide ions:
[OH-] = 10^-(10.1)
[OH-] = 7.943 x 10^(-11)
2. Calculate the moles of HNO3:
[H3O+] = [OH-] = 7.943 x 10^(-11) M
moles of HNO3 = [H3O+] x 0.59
moles of HNO3 = 7.943 x 10^(-11) x 0.59
3. Calculate the grams of HNO3:
grams of HNO3 = moles of HNO3 x molar mass of HNO3
grams of HNO3 = (7.943 x 10^(-11) x 0.59) x 63.01
After performing the calculations, the resulting grams of HNO3 is approximately 2.9 x 10^(-9) g.
Therefore, the correct answer is (a) 2.9 x 10^(-9) g.
To solve this problem, we need to consider the relationship between pOH and the concentration of hydroxide ions (OH-) in a solution.
The pOH is defined as the negative logarithm (base 10) of the hydroxide ion concentration. The formula for pOH is given by the expression:
pOH = -log[OH-]
In this case, we are given a pOH value of 10.1. To find the concentration of hydroxide ions, we need to convert the pOH into a hydronium ion concentration (H+). Since water is neutral, the concentration of hydroxide ions is equal to the concentration of hydronium ions in a neutral solution, which is 1 x 10^-7 M.
Next, we need to use the balanced chemical equation for the ionization of nitric acid (HNO3) in water to find the molar ratio between HNO3 and OH-. The balanced equation is:
HNO3 + H2O -> H3O+ + NO3-
From this equation, we can see that for every 1 mole of HNO3, we get 1 mole of H3O+ ions. The concentration of H3O+ ions is therefore equal to the concentration of HNO3 that needs to be dissolved in water to achieve the desired pOH value.
Now we have all the necessary information to calculate the amount of HNO3 needed. We know the concentration of H3O+ ions is 1 x 10^-7 M, and we have a volume of 590 ml of water.
First, we need to convert the volume from milliliters to liters:
590 ml = 0.59 L
Next, we can use the formula:
concentration (M) = moles/volume (L)
Rearranging the formula gives:
moles = concentration (M) x volume (L)
moles = (1 x 10^-7 M) x (0.59 L)
moles = 5.9 x 10^-8 mol
Finally, we can convert the moles of HNO3 into grams using the molar mass of HNO3, which is approximately 63 g/mol.
grams = moles x molar mass
grams = (5.9 x 10^-8 mol) x (63 g/mol)
grams = 3.717 x 10^-6 g
Therefore, the answer is d. 7.9 x 10^-6 g (rounded to three significant figures).