QuestionJanuary 7, 2026

The formula for pH is pH=-log[H^+] where [H^+] is the hydrogen ion concentration. The hydrogen lon concentration, [H^+] In a certain brand of fresh-brewed coffee is [H^+]=1.3times 10^-5 . What is the pH of the product, to the nearest tenth? The pH of the coffee is square (Round to the nearest tenth as needed.)

The formula for pH is pH=-log[H^+] where [H^+] is the hydrogen ion concentration. The hydrogen lon concentration, [H^+] In a certain brand of fresh-brewed coffee is [H^+]=1.3times 10^-5 . What is the pH of the product, to the nearest tenth? The pH of the coffee is square (Round to the nearest tenth as needed.)
The formula for pH is pH=-log[H^+] where [H^+] is the hydrogen ion concentration. The hydrogen lon concentration, [H^+] In a certain brand of fresh-brewed coffee is [H^+]=1.3times 10^-5 . What is the pH of the product, to the nearest tenth? The pH of the coffee is square (Round to the nearest tenth as needed.)

Solution
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Answer

4.9 Explanation 1. Write the pH formula The formula is **pH = -\log [H^+]**. 2. Substitute the given concentration Given [H^+] = 1.3 \times 10^{-5}, substitute into the formula: pH = -\log(1.3 \times 10^{-5}). 3. Simplify using log properties \log(1.3 \times 10^{-5}) = \log(1.3) + \log(10^{-5}) = \log(1.3) - 5. So, pH = -(\log(1.3) - 5) = 5 - \log(1.3). 4. Compute numeric value \log(1.3) \approx 0.1139. Thus, pH = 5 - 0.1139 = 4.8861. 5. Round to the nearest tenth pH \approx 4.9.

Explanation

1. Write the pH formula <br /> The formula is **$pH = -\log [H^+]$**.<br /><br />2. Substitute the given concentration <br /> Given $[H^+] = 1.3 \times 10^{-5}$, substitute into the formula: <br />$pH = -\log(1.3 \times 10^{-5})$.<br /><br />3. Simplify using log properties <br /> $\log(1.3 \times 10^{-5}) = \log(1.3) + \log(10^{-5}) = \log(1.3) - 5$. <br />So, $pH = -(\log(1.3) - 5) = 5 - \log(1.3)$.<br /><br />4. Compute numeric value <br /> $\log(1.3) \approx 0.1139$. <br />Thus, $pH = 5 - 0.1139 = 4.8861$.<br /><br />5. Round to the nearest tenth <br /> $pH \approx 4.9$.
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