# what is the ph of a solution with [h3o ] = 1 × 10-9 m?

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What is the pH of a solution with H3O+ 1 10 9 m?, The pH of a solution with a concentration of H3 O+ of 1×10−9 M is 9.

#### Explanation:

To calculate the pH of a solution, use the equation

pH=log[H3O+]

Since we are given the hydronium ion (H3O+) concentration of the solution, we can plug this value into the formula.

pH=log[1.00×109M]
pH=9

The pH of this solution is 9.

pH=pH3O+ because hydrogen ions attach to water molecules for form hydronium ions.Some other key things to remember for pH calculations include:pH + pOH = 14

pH = -log[H+]
pOH = -log[OH]

[H+]=10pH
[OH]=10pOH

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## How do you determine pH of a solution?

pH is defined as
pH = -log₁₀[H₃O⁺]

There are five possible scenarios:
You are given the concentration of H₃O⁺.
You are given the concentration of a strong acid.
You are given the concentration and the Ka of a weak acid that is less than 5 % dissociated.
You are given the concentration and the Ka of a weak acid that is more than 5 % dissociated.
You are given or have calculated the OH⁻ concentration.

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Let’s look at each of these in turn.

Calculate pH of a solution in which [H₃O⁺] = 1.4 × 10⁻⁵ mol/L
pH = -log₁₀[H₃O⁺]
pH = -log10(1.4 × 10⁻⁵)
pH = 4.85

Find the pH of a 0.0025 mol/L HCl solution. The HCl is a strong acid and is 100% ionized in water. The hydronium ion concentration is 0.0025 mol/L. Thus:
pH = -log₁₀[H₃O⁺] = -log10(0.0025) = 2.60Calculate the pH of a 0.100 mol/L solution of acetic acid in water. To do this, we first write down the equilibrium involved and the expression for the equilibrium constant. We’ll use HA to represent the acetic acid.

HA + H₂O → H₃O⁺ + A⁻
Ka = [H₃O⁺][ A⁻]/[HA] = 1.76 ×10⁻⁵

Under the chemical equation we will write a so-called ICE table, in which we list the Initial, Change, and Final concentrations. We use this just as a way to keep our numbers organized.

HA  +   H₂O →   H₃O⁺    +   A⁻


I: 0.100 – –
C: -x +x +x
E: 0.100 – x x x

(Sorry, this editor won’t allow me to line up the entries properly under each other.)
The initial concentration of HA is 0.100 mol/L, and the concentrations of H₃O⁺ and A⁻ are negligible. We don’t know how much of the acetic acid ionizes, but we can say that its concentration decreases by x mol/L, while the concentrations of H₃O⁺ and A⁻ increase by the same amount. This gives us the equilibrium concentrations, which we can now insert into the Ka expression.

Ka = [H₃O⁺][ A⁻]/[HA] = x•x/(0.100 – x) = x²/(0.100 – x) = 1.76 ×10⁻⁵

If we try to solve this for x, we will get an equation with terms in x², x, and a constant, i.e., a quadratic equation. Fortunately, if x is small enough, then
0.100 – x ~ 0.100, and we will then have a perfect square, which is much easier to solve.

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How can we tell if x is negligible? A common test is: if the initial concentration of HA divided by Ka is greater than 400, then x is negligible.

0.100/1.76 ×10⁻⁵ = 5680. This is greater than 400, so we can ignore the x in the denominator.

x²/0.100 = 1.76 ×10⁻⁵
x² = 0.100 × 1.76 × 10⁻⁵ = 1.76 × 10⁻⁶
x = √(1.76 × 10⁻⁶) = 1.33 × 10⁻³
[H₃O⁺] = 1.33 × 10⁻³ mol/L
pH = -log₁₀[H₃O⁺] = -log10(1.33 × 10⁻³) = 2.88

Calculate the pH of a 0.0010 mol/L solution of acetic acid in water. As before, we first write down the equilibrium involved and the expression for the equilibrium constant, with an ICE table in between.

HA  +   H₂O →   H₃O⁺    +   A⁻


I: 0.00100 – –
C: -x +x +x
E: 0.00100 – x x x
Ka = [H₃O⁺][ A⁻]/[HA] = x•x/(0.00100 – x) = x²/(0.00100 – x) = 1.76 ×10⁻⁵

Now we test to see whether x is negligible.
0.00100/1.76 ×10⁻⁵ = 56.8. This is less than 400, so have no choice but to solve a quadratic equation.x² = (0.00100 – x) × 1.76 × 10⁻⁵ = 1.76 × 10⁻⁷ – 1.76 × 10⁻⁵x
x² + 1.76 ×10⁻⁵x – 1.76 × 10⁻⁷ = 0We use the quadratic equation formula x = [-b ± √(b² – 4ac)]/2a to get
x = 4.11 × 10⁻⁴
[H₃O⁺] = 4.11 × 10⁻⁴ mol/L
pH = -log₁₀[H₃O⁺] = -log₁₀(4.11 × 10⁻⁴) = 3.39

Finally, what is the pH of a solution in which the concentration of OH⁻ is
0.0500 mol/L.
Here, it is easier to calculate the pOH and then calculate the pH.
The relationships are
pOH = -log₁₀[OH⁻] and pH + pOH = 14.00
pOH = -log₁₀[OH⁻] = -log₁₀(0.0500) = 1.30
pH = 14.00 – pOH = 14.00 – 1.30 = 12.70