What is the [.H+]. in a 0.40 M solution of HOCl, Ka=3.5× 10- 8 ?

Question

What is the [.H+]. in a 0.40 M solution of HOCl, Ka=3.5× 10- 8 ? (A) 1.4× 10- 8 M (B) 1.2× 10- 4 M (C) 1.9× 10- 4 M (D) 3.7× 10- 4 M

Tardigrade Admin · Accepted Answer

textHOCl oversetleftharpoons text textH text++ textOCl- C(1 - α ) textCα Cα textK texta = textCα2 textH⊕ = textCα = √ textK texta × textC =√3.5 × 10- 8 × 0.40 [ textH text+]=1.183× 10- 4

Q. What is the $[H^{+}]$ in a 0.40 M solution of HOCl, $K_{a} = 3.5 \times 1 0^{- 8}$ ?

$1.4 \times 1 0^{- 8} M$

$1.2 \times 1 0^{- 4} M$

$1.9 \times 1 0^{- 4} M$

$3.7 \times 1 0^{- 4} M$

$HOCl ⇌ H^{+} + OCl^{-}$
$C (1 - α) C α C α$
$K_{a} = Cα^{2}$
$H^{\oplus} = Cα = K_{a} \times C$
$= 3.5 \times 1 0^{- 8} \times 0.40$
$[H^{+}] = 1.183 \times 1 0^{- 4}$

Q. What is the [H+] in a 0.40 M solution of HOCl, Ka​=3.5×10−8 ?

1.4×10−8M

1.2×10−4M

1.9×10−4M

3.7×10−4M