A solution is prepared by adding 47.3 mL of concentrated hydrochloric acid and 16.3 mL of concentrated nitric acid to 300 mL of water. More water is added until the final volume is 1.00 L. Calculate [H+], [OH -], and the pH for this solution. [Hint: Concentrated HCl is 38% HCl (by mass) and has a density of 1.19 g/mL; concentrated HNO3 is 70.% HNO3 and has a density of 1.42 g/mL.]
Find [H+] [OH-] and Ph
The easiest way to explain this (but perhaps the long way around) is to first convert to molarity for both HCl and HNO3.
For HCl:
1.19 g/mL x 1000 mL x 0.38 x (1 mol/molar mass HCl) = mols HCl/L = M
For HNO3:
density g/mL x 1000 mL x 0.70 x (1 mol/molar mass HNO3) = moles HNO3/L = M
Then mols HCl = M x L = M x 0.0473 = ?
mols HNO3 = M x 0.0163 = ?
Add mols HCl to mol HNO3.
M soln = total mols/total volume (which is 1.0 L final volume).
Note: Those aren't hints at the end of the problem. That information is essential to the problem.)
To calculate the [H+], [OH-], and pH for this solution, we need to determine the amount of each acid added and then use that information to calculate the concentration of H+ ions.
Step 1: Calculate the amount of HCl added
The volume of concentrated HCl added is 47.3 mL. However, we need to find the mass of HCl added. Since the concentration of HCl is given as a percentage by mass and the density is provided, we can calculate the mass using the following formula:
mass = volume x density
mass of HCl = 47.3 mL x 1.19 g/mL
mass of HCl = 56.287 g
Step 2: Calculate the moles of HCl added
To calculate the moles of HCl added, we need to use the molar mass of HCl. The molar mass of HCl is 36.46 g/mol. We can use the following formula:
moles of HCl = mass of HCl / molar mass
moles of HCl = 56.287 g / 36.46 g/mol
moles of HCl = 1.541 mol
Step 3: Repeat steps 1 and 2 for HNO3
Using the same approach as for HCl, we can calculate the moles of HNO3 added. The mass of HNO3 can be calculated using the volume and density given:
mass of HNO3 = 16.3 mL x 1.42 g/mL
mass of HNO3 = 23.266 g
The molar mass of HNO3 is 63.01 g/mol:
moles of HNO3 = 23.266 g / 63.01 g/mol
moles of HNO3 = 0.3691 mol
Step 4: Calculate the total moles of H+ ions
To find the total moles of H+ ions, we add the moles of HCl and HNO3 together:
moles of H+ = moles of HCl + moles of HNO3
moles of H+ = 1.541 mol + 0.3691 mol
moles of H+ = 1.9101 mol
Step 5: Calculate the concentration of H+ ions
The final volume of the solution is 1.00 L. Therefore, the concentration of H+ ions can be calculated using the moles of H+ ions and the volume:
[H+] = moles of H+ / volume
[H+] = 1.9101 mol / 1.00 L
[H+] = 1.9101 M
Step 6: Calculate [OH-]
In neutral water, [H+] and [OH-] are equal. Since this is not a neutral solution, [OH-] can be calculated using the following equation:
[H+] x [OH-] = 1.0 x 10^-14 M^2
[OH-] = 1.0 x 10^-14 M^2 / [H+]
[OH-] = 1.0 x 10^-14 M^2 / 1.9101 M
[OH-] = 5.23 x 10^-15 M
Step 7: Calculate pH
pH is a measure of the acidity of a solution and is defined as the negative logarithm (base 10) of the concentration of H+ ions:
pH = -log[H+]
pH = -log(1.9101)
pH = 0.718
Therefore, in this solution, the [H+] is 1.9101 M, the [OH-] is 5.23 x 10^-15 M, and the pH is 0.718.