Calculate the pH of a 0.10 m solution of hypochlorous acid, hocl. ka of hocl is 3.5×10−8 at 25 ∘c. express your answer numerically using two decimal places.
4.23. Explanation: ∵ pH = – log[H⁺]. For weak acids: ∵ [H⁺] = √(ka)(c). ∴ [H⁺] = √(3.5 × 10⁻⁸)(0.10 M) = 5.92 × 10⁻⁵. ∴ pH = – log[H⁺] = – log(5.92 x 10⁻⁵) = 4.2279 ≅ 4.23.
1 Explanation: First we need to know the chemical formula of hypochlorous acid. It’s HOCl. The ionization equation is: HOCl ⇌ ClO− + H This means that in each mole of acid, exactly one mole of hydrogen ions is produced. This also means that there is exactly 0.1 M H+ in 0.1 M acid. Now let’s calculate the pH of the acidic solution. By definition, pH is the negative logarithm to base 10 of the hydrogen ion concentration of the acid. Mathematically: pH = -log[H+] now we know that [H+] = 0.1 M pH = -log[0.1] pH = 1 The pH of 0.1 M hypochlorous acid is 1
we know this:
Ka = [H+][OCl-] /[HOCl]
and HOCl dissociates into an equal number of H+ = OCl- ions
we replace [OCl-] by [H+}
Hence
Ka = [H+][H+] /{HOCl]
Ka = [H+]^2 / [HOCl]
[H+]^2 = Ka X [HOCl]
[H+] = sqrt{Ka X [HOCl]}
[H+] = square{ 3.5 x 10^-8 x 0.10}
[H+] = square[ 3.5 x 10^-9
[H+] = 5.916 x 10^-5
you know: pH = -log(10) [H+]
pH = -log(10) 5.916 x 10^-5
pH = -(-4.227965)
pH = 4.227965
pH=4.23.