A solution is prepared using 15.0 mL of 1.0 M HCl and 20.0 mL of 0.5 M HNO3. The final volume of the solution is 1.25 L. What are the [H3O+] and [OH-] in the final solution? What is the pH of the final solution?

Question

A solution is prepared using 15.0 mL of 1.0 M HCl and 20.0 mL of 0.5 M HNO3. The final volume of the solution is 1.25 L. What are the [H3O+] and [OH-] in the final solution? What is the pH of the final solution?

Answer 1

moles HCl = M x L

moles HNO3 M x L
M final solution = total moles/total L.

Answer 2

Ok. I got 0.015 mol HCl and 0.1 mol HNO3. Do I just add the moles to get the total? So 0.115 total moles. Which would make a the Molarity of the final solution 0.92M?

Answer 3

I mean 0.092M.

Answer 4

To find the [H3O+] and [OH-] in the final solution, we need to consider the reaction between HCl and HNO3. These are both strong acids and completely dissociate in water.

The balanced equation for the reaction is:

HCl + HNO3 → H2O + Cl- + NO3-

Since HCl is a stronger acid than HNO3, it will donate its proton (H+) to HNO3, resulting in the formation of water and the chloride (Cl-) and nitrate (NO3-) ions.

To calculate the concentration of HCl and HNO3 after mixing, we can use the principle of moles and volumes:

moles of HCl = concentration (1.0 M) * volume (15.0 mL)
moles of HNO3 = concentration (0.5 M) * volume (20.0 mL)

Next, we need to convert the volumes of HCl and HNO3 to liters:

volume (HCl) = 15.0 mL = 15.0 mL / 1000 mL/L = 0.015 L
volume (HNO3) = 20.0 mL = 20.0 mL / 1000 mL/L = 0.020 L

Now we can calculate the moles of HCl and HNO3:

moles of HCl = 1.0 M * 0.015 L = 0.015 mol
moles of HNO3 = 0.5 M * 0.020 L = 0.010 mol

Since both HCl and HNO3 are strong acids, they will completely dissociate, resulting in the formation of equal moles of H3O+ (hydronium ions) and the respective anions (Cl- and NO3-).

Therefore, the final concentration of H3O+ in the solution is:

[H3O+] = moles of H3O+ / volume of solution

moles of H3O+ = moles of HCl + moles of HNO3 = 0.015 mol + 0.010 mol = 0.025 mol
volume of solution = 1.25 L

[H3O+] = 0.025 mol / 1.25 L = 0.020 M

Since water is neutral and contains equal concentrations of H3O+ and OH-, we can calculate the [OH-] in the final solution:

[OH-] = 1.0 x 10^-14 / [H3O+]
[OH-] = 1.0 x 10^-14 / 0.020 M = 5.0 x 10^-13 M

To find the pH of the final solution, we can use the formula:

pH = -log[H3O+]

pH = -log(0.020) ≈ 1.70

Therefore, the [H3O+] in the final solution is 0.020 M, the [OH-] is 5.0 x 10^-13 M, and the pH is approximately 1.70.