Water in equilibrium with air contains . The resulting carbonic acid, , gives the solution a hydroni — Ionic Equilibrium Chemistry Question
Question
Water in equilibrium with air contains $4.4 \times 10^{-5}\% \text{ CO}_2$. The resulting carbonic acid, $\text{H}_2\text{CO}_3$, gives the solution a hydronium ion concentration of $2.0 \times 10^{-6} \text{ M}$, about 20 times greater than that of pure water. What is the pH of the solution at 298 K? ($\log 4.4 = 0.64$, $\log 2 = 0.3$)
Answer: B
💡 Solution & Explanation
The hydronium ion concentration is directly given as $[\text{H}_3\text{O}^+] = 2.0 \times 10^{-6} \text{ M}$. The $\text{pH}$ can be calculated directly from this concentration: $\text{pH} = -\log(2.0 \times 10^{-6}) = 6 - \log 2$. Given $\log 2 = 0.3$, $\text{pH} = 6 - 0.3 = 5.70$. Therefore, correct answer is B.
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