a solution is prepared using .015 L of 1.0 M HCl and .020 L of .50 M HNO3. the final volume of the solution is 1.25 L. Answer th following questions.

a)what are the [H3O+] and [OH-] in the final solution?
b)What is the pH of the final solution?

moles HCl = M x L

moles HNO3 = M x L
M final soln = total moles/total liters.
(H^+) = molarity
Then pH = -log(H^+)

Thanks!

To find the [H3O+] and [OH-] in the final solution, we need to calculate the amount of H3O+ and OH- ions produced by the given acids.

a) First, let's calculate the moles of H3O+ and OH- ions produced by 1.0 M HCl.

Moles of H3O+ from HCl = volume of HCl (in liters) * concentration of HCl
= 0.015 L * 1.0 mol/L
= 0.015 mol

Since HCl is a strong acid and dissociates completely, the number of moles of H3O+ ions produced is the same as the number of moles of HCl used.

Next, let's calculate the moles of H3O+ and OH- ions produced by 0.50 M HNO3.

Moles of H3O+ from HNO3 = volume of HNO3 (in liters) * concentration of HNO3
= 0.020 L * 0.50 mol/L
= 0.010 mol

Since HNO3 is also a strong acid, the moles of H3O+ ions produced are the same as the moles of HNO3 used.

Now, let's calculate the total moles of H3O+ ions in the final solution.

Total moles of H3O+ = moles of H3O+ from HCl + moles of H3O+ from HNO3
= 0.015 mol + 0.010 mol
= 0.025 mol

Since the volume of the final solution is 1.25 L, we can find the concentration ([H3O+]).

[H3O+] = moles of H3O+ ions / volume of the solution
= 0.025 mol / 1.25 L
= 0.02 mol/L

Since the concentration of [OH-] can be calculated using the equation for water dissociation constant (Kw), where Kw = [H3O+][OH-] = 1.0 x 10^-14 at 25 degrees Celsius, we can calculate [OH-].

[OH-] = Kw / [H3O+]
= (1.0 x 10^-14) / (0.02 mol/L)
= 5.0 x 10^-13 mol/L

Therefore, in the final solution, the concentrations of [H3O+] and [OH-] are 0.02 mol/L and 5.0 x 10^-13 mol/L, respectively.

b) To find the pH of the final solution, we can use the equation:

pH = -log[H3O+]

Using the value of [H3O+] calculated above, we can compute the pH.

pH = -log(0.02)
= 1.7

So, the pH of the final solution is 1.7.