Question

use the references to access important values if needed for thi

the ph of an aqueous solution of 0.1650 m hydrosulfuric acid, h₂s (aq), is

submit answer retry entire group 6 more group attempts remaining

Explanation:

Step1: Consider the first dissociation of \(H_2S\)

\(H_2S(aq)
ightleftharpoons H^+(aq)+HS^-(aq)\), \(K_{a1}=8.9\times 10^{-8}\) (from references). Since \(K_{a1}\ll1\), we can use the approximation \( [H^+]=\sqrt{K_{a1}\times c}\) where \(c = 0.1650M\)

$$ [H^+]=\sqrt{8.9\times 10^{-8}\times0.1650}$$

$$ [H^+]=\sqrt{1.4685\times 10^{-8}}$$

$$ [H^+]=1.21\times 10^{-4}M$$

Step2: Calculate the pH

The formula for pH is \(pH=-\log[H^+]\)

$$pH =-\log(1.21\times 10^{-4})$$

$$pH=-( \log(1.21)+\log(10^{-4}))$$

$$pH=-(0.083 - 4)$$

$$pH = 3.917\approx3.92$$

Answer:

\(3.92\)