Gise3lchelorn
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What is the [h3o ] in a buffer that consists of 0.30 m hcooh and 0.20 m hcoona? ka = 1.7 × 10-4 i just need the steps to do it, not the answer ?

Respuesta :

Yipes
HA + H₂O ⇔ H₃O⁺ + A⁻

[tex]K_{k} = \frac{|H_{3}O^{+}||A^{-}|}{|HA|}\\\\ |H_{3}O^{+}|=K_{k}*\frac{|HX|}{|A^{-}|}\\\\ |HX| \sim C_{k}\\\\ |A^{-}| \sim C_{s}\\\\ |H_{3}O^{+}|=K_{k}*\frac{C_{k}}{C_{s}}[/tex]

Why?

HCOONa --> strong base and weak acid, α=100% - [HCOO⁻] ~ Cs
HCOOH --> weak acid α<100% - [H⁺] ~ Ck

[tex]|H_{3}O^{+}|=K_{k}*\frac{C_{k}}{C_{s}}=1,7*10^{-4}*\frac{0,3}{0,2}=2,55*10^{-4}M[/tex]
dsdrajlin

The concentration of H₃O⁺ in a buffer system formed by 0.30 M HCOOH and 0.20 M HCOONa is 2.5 × 10⁻⁴ M.

We have a buffer system formed by 0.30 M HCOOH and 0.20 M HCOONa.

We can calculate the pH of the buffer using Henderson-Hasselbach's equation.

[tex]pH = pKa + log \frac{[HCOONa]}{[HCOOH]} \\\\pH = -log (1.7 \times 10^{-4} )+ log \frac{0.20}{0.30} = 3.6[/tex]

Given the pH is 3.6, we can calculate the concentration of H₃O⁺ using the following expression.

[tex]pH = -log [H_3O^{+} ]\\[H_3O^{+} ] = antilog-pH = antilog -3.6 = 2.5 \times 10^{-4} M[/tex]

The concentration of H₃O⁺ in a buffer system formed by 0.30 M HCOOH and 0.20 M HCOONa is 2.5 × 10⁻⁴ M.

Learn more: https://brainly.com/question/13564840

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