Question

Determine whether the following hydroxide ion concentrations ([OH−]) correspond to acidic, basic, or neutral solutions by estimating their corresponding hydronium ion concentrations ([H3O+] using the ion product constant of water (Kw).

Hydronium ion concentration [H3O+] Solution condition
Greater than 1×10−7 M Acidic
Equal to 1×10−7 M Neutral
Less than 1×10−7 M Basic
It may help to keep the following equation in mind as you work

Kw = [H3O+][OH−] = [1×10−7 M][1×10−7 M] = 1×10−14 M

Drag the appropriate items to their respective bins.

Hints

[OH−] = 6×10−12 M

[OH−] = 9×10−9 M

[OH−] = 8×10−10 M

[OH−] = 7×10−13 M

[OH−] = 2×10−2 M

[OH−] = 9×10−4 M

[OH−] = 5×10−5 M

[OH−] = 1×10−7 M

Acidic

Neutral

Basic

Part B

A solution has [H3O+] = 5.2×10−5M . Use the ion product constant of water

Kw=[H3O+][OH−]

to find the [OH−] of the solution.

Part C

A solution has [OH−] = 2.7×10−2M . Use the ion product constant of water

Kw=[H3O+][OH−]

to find the [H3O+] of the solution.

Answer 1

See explanation below

Step-by-step explanation:

To do this, we will use the following expression to calculate the [H⁺]:

[H⁺] = Kw / [OH⁻]

[H⁺] is the same as [H₃O⁺]. So we have the [OH⁻] so, let's replace every value into the above expression to calculate the hydronium concentration and say if it's acidic, basic or neutral. This can be known because if the [H⁺] > 1x10⁻⁷ M the solution is acidic. If it's [H⁺] < 1x10⁻⁷ M the solution is basic, and if it's [H⁺] = 1x10⁻⁷ M the solution is neutral.

a) [H⁺] = 1x10⁻¹⁴ / 6x10⁻¹² = 1.67x10⁻³ M. Acidic.

b) [H⁺] = 1x10⁻¹⁴ / 9x10⁻⁹ = 1.11x10⁻⁶ M. Acidic.

c) [H⁺] = 1x10⁻¹⁴ / 8x10⁻¹⁰ = 1.25x10⁻⁵ M. Acidic.

d) [H⁺] = 1x10⁻¹⁴ / 7x10⁻¹³ = 0.0143 M. Acidic.

e) [H⁺] = 1x10⁻¹⁴ / 2x10⁻² = 5x10⁻¹³ M. Basic.

f) [H⁺] = 1x10⁻¹⁴ / 9x10⁻⁴ = 1.11x10⁻¹¹ M. Basic.

g) [H⁺] = 1x10⁻¹⁴ / 5x10⁻⁵ = 2x10⁻¹⁰ M. Basic.

h) [H⁺] = 1x10⁻¹⁴ / 1x10⁻⁷ = 1x10⁻⁷ M. Neutral.

Part B.

In this part, we'll use the following expression and replace the given values:

[OH⁻] = Kw / [H⁺]

Replacing the values:

[OH⁻] = 1x10⁻¹⁴ / 5.2x10⁻⁵

[OH⁻] = 1.92x10⁻¹⁰ M

PArt C:

In this case, we will use expression of part A, and replace the given values:

[H⁺] = 1x10⁻¹⁴ / 2.7x10⁻²

[H⁺] = 3.7x10⁻¹³ M