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Calculate the pH, [H₃O+], [OH-], and pOH of 500.0 mL of 0.09933 M HNO₃ solution.

a) pH:
b) [H₃O+]:
c) [OH-]:
d) pOH:

User Maltronic
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1 Answer

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Final answer:

For a 0.09933 M HNO3 solution, the concentration of H3O+ is equivalent to the acid concentration (0.09933 M), leading to a pH of 1.003. The [OH-] is calculated using the ion-product constant for water and is approximately 1.007 × 10⁻¹³ M with a corresponding pOH of 12.997.

Step-by-step explanation:

We are dealing with a 0.09933 M HNO₃ solution, which is a strong acid and will dissociate completely in water to produce H₃O⁺ and NO₃⁻ ions. The concentration of H₃O⁺ will be the same as the initial concentration of the acid, so [H₃O⁺] = 0.09933 M. To calculate the pH, we take the negative logarithm (base 10) of the hydrogen ion concentration:
pH = -log(0.09933) = 1.003. Now, let's find the concentration of OH⁻ ions. Since H₃O⁺ and OH⁻ are inversely related by the equation Kw = [H₃O⁺][OH⁻], where Kw is the ion-product constant for water at 25 °C (1.0 × 10⁻¹⁴), we have [OH⁻] = Kw / [H₃O⁺] = 1.0 × 10⁻¹⁴ / 0.09933 M ≈ 1.007 × 10⁻¹³ M. The pOH can be calculated as pOH = -log([OH⁻]) = -log(1.007 × 10⁻¹³) ≈ 12.997. Final results are:

  • pH = 1.003
  • [H₃O⁺] = 0.09933 M
  • [OH⁻] = 1.007 × 10⁻¹³ M
  • pOH = 12.997

User Yiu
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