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Calculus IChapter 1: ReviewSection 1-8

Logarithm Functions — Question 2

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Question 2

A scientist is studying the pH level of a solution, which is defined by the formula: pH=−log⁡10[H+]\text{pH} = -\log_{10}[H^+] where [H+][H^+] is the concentration of hydrogen ions in moles per liter.

  • (a) If a solution has a hydrogen ion concentration of [H+]=3.2×10−5[H^+] = 3.2 \times 10^{-5}, find its pH.

  • (b) If another solution has a pH of 4.6, what is its hydrogen ion concentration?

  • (c) Which solution is more acidic, and by how many times is it more acidic?

Original worksheet page 1: question and worked solution for 1-8-002
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Question 2 - Solution

(a) Compute pH: pH=−log⁡10(3.2×10−5)=−[log⁡10(3.2)+log⁡10(10−5)]\text{pH} = -\log_{10}(3.2 \times 10^{-5}) = -[\log_{10}(3.2) + \log_{10}(10^{-5})] =−[log⁡10(3.2)−5]≈−[0.5051−5]=−(−4.4949)=4.49= -[\log_{10}(3.2) - 5] \approx -[0.5051 - 5] = -(-4.4949) = \boxed{4.49}

(b) Compute hydrogen ion concentration from pH: pH=−log⁡10[H+]⇒log⁡10[H+]=−4.6⇒[H+]=10−4.6≈2.51×10−5\text{pH} = -\log_{10}[H^+] \Rightarrow \log_{10}[H^+] = -4.6 \Rightarrow [H^+] = 10^{-4.6} \approx \boxed{2.51 \times 10^{-5}}

(c) Compare acidity:

Recall: lower pH = more acidic.

The first solution has pH ≈4.49\approx 4.49, second has pH 4.64.6, so first solution is more acidic.

To find how many times more acidic: [H+]first[H+]second=3.2×10−52.51×10−5≈1.27 times more acidic\frac{[H^+]_{\text{first}}}{[H^+]_{\text{second}}} = \frac{3.2 \times 10^{-5}}{2.51 \times 10^{-5}} \approx \boxed{1.27 \text{ times more acidic}}

Original worksheet page 2: question and worked solution for 1-8-002

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