For each strong acid solution, determine [H3O ], [OH-], and pH. a. 0.25 M HCl b. 0.015 M HNO3 c. a solution that is 0.052 M in HBr and 0.020 M in HNO3 d. a solution that is 0.655% HNO3 by mass (assume a density of 1.01 g>mL for the solution)

See explanation below

Explanation:

First to all, we know that all of these acids are strong acid, this means that when they are in solution, they will dissociate completely in solution. So, to get the [H₃O⁺] is basically the same concentration of the original acid.

To get the [OH⁻], we just use the expression for the constant dissociation of water (Kw = 1x10⁻¹⁴) and solve for the [OH⁻]:

Kw = [H₃O⁺] [OH⁻] > [OH⁻] = Kw/[H₃O⁺]   (1)

To get the pH we use the following expression:

pH = -log[H₃O⁺]   (2)

Using these, let's solve the first three exercises:

a) [H₃O⁺] = [HCl] = 0.25 M

[OH⁻] = 1x10⁻¹⁴ / 0.25 = 4x10⁻¹⁴ M

pH = -log(0.25) = 0.60

b) [H₃O⁺] = [HNO₃] = 0.015 M

[OH⁻] = 1x10⁻¹⁴ / 0.015 = 6.67x10⁻¹³ M

pH = -log(0.015) = 1.82

c) In this case we have two acids in solution, so the concentration of the solution will be the sum of the two acids so:

[H₃O⁺] = [HBr] + [HNO₃] = 0.02 + 0.052 = 0.072 M

[OH⁻] = 1x10⁻¹⁴ / 0.072 = 1.39x10⁻¹³ M

pH = -log(0.072) = 1.14

d) in this case, we have the %m/m of the acid and it's density, so the concentration in molarity can be calculated using the following expression:

M = d * % * 1000 / MM * 10

the molecular mass of HNO₃  is 63 g/mol, so the concentration of the acid is:

[HNO₃] = [H₃O⁺] = 0.655 * 1.01 * 1000 / 63 * 100 = 0.105 M

[OH⁻] = 1x10⁻¹⁴ / 0.105 = 9.52x10⁻¹⁴ M

pH = -log(0.105) = 0.98 M

And all of this, will be the correct answer

d: the voltage of the secondary coil is less than that of the primary coil.

im not suree sorry if wrong

practical process described pest part of scientific investigation.

because practical knowledge is powerful then therotical knowledge.