For the strong acids find the [H3O]and [OH] in a solution that is .785% HNO3 by mass (assume a density of 1.01 g/ml). If someone could offer step by step instructions, that would be terrific!
0.785% w/w means 0.785 g HNO3/100 g soln.
Density = 1.01 g/mL; therefore, the 100 g soln has a volume of 99 mL.
0.785 g is how many mols? That's 0.785/63 = approximately 0.012 and the M = mols/L = 0.012 mols/0.099 L = approximately 0.13M. (You need to redo all of this and use better numbers).
(H3O^+) = 0.13M
(H3O^+)(OH^-) = Kw = 1E-14
You know H3O^+ and Kw, solve for OH^-
Thank you so much for your help! I was able to follow your steps very well! Much appreciated DrBob222!
To find the [H3O+] and [OH-] in a solution of HNO3, we need to follow these steps:
Step 1: Convert the mass percent of HNO3 to grams.
Given that the solution is 0.785% HNO3 by mass and has a density of 1.01 g/ml, we can start by assuming we have 100 grams of the solution (since percent mass is always measured per 100 grams).
Therefore, the mass of HNO3 in the solution is:
(0.785/100) * 100 grams = 0.785 grams of HNO3
Step 2: Calculate the moles of HNO3.
We need to convert the mass of HNO3 to moles using its molar mass.
The molar mass of HNO3 is:
1*1 (H) + 1*14 (N) + 3*16 (O) = 1 + 14 + 48 = 63 g/mol
Thus, the number of moles of HNO3 is:
0.785 grams / 63 g/mol ≈ 0.012 moles of HNO3
Step 3: Use stoichiometry to determine the concentrations of [H3O+] and [NO3-].
For every 1 mole of HNO3 that dissociates in water, it produces 1 mole of H3O+ and 1 mole of NO3-.
Therefore, the concentrations of [H3O+] and [NO3-] in the solution are:
[H3O+] = 0.012 moles / 0.1 L (assuming we have 100 mL of the solution) = 0.12 M
[NO3-] = 0.012 moles / 0.1 L = 0.12 M
Step 4: Find the concentration of [OH-].
For a strong acid like HNO3, it fully dissociates in water, so the concentration of [OH-] will be negligible in this case.
Therefore, [OH-] ≈ 0 M.
In summary, for a solution containing 0.785% HNO3 by mass, the concentration of [H3O+] is 0.12 M, and the concentration of [OH-] is approximately 0 M.