Question

the hydrogen ion concentration of a sample of each substance is given. calculate the ph of the substance. (round your answers to one decimal place.) (a) oranges: h⁺ = 2.5 × 10⁻⁴ m (b) tomato juice: h⁺ = 3.2 × 10⁻⁴ m (c) seawater: h⁺ = 5.0 × 10⁻⁹ m

Explanation:

Step1: Recall the pH formula

The formula to calculate pH is \( \text{pH} = -\log_{10}[\text{H}^+] \), where \([\text{H}^+]\) is the hydrogen ion concentration in moles per liter (M).

Step2: Calculate pH for Oranges

For oranges, \([\text{H}^+] = 2.5 \times 10^{-4}\) M.
Using the pH formula:
\( \text{pH} = -\log_{10}(2.5 \times 10^{-4}) \)
First, calculate \( \log_{10}(2.5 \times 10^{-4}) = \log_{10}(2.5) + \log_{10}(10^{-4}) \)
We know that \( \log_{10}(2.5) \approx 0.3979 \) and \( \log_{10}(10^{-4}) = -4 \)
So, \( \log_{10}(2.5 \times 10^{-4}) \approx 0.3979 - 4 = -3.6021 \)
Then, \( \text{pH} = -(-3.6021) \approx 3.6 \) (rounded to one decimal place)

Step3: Calculate pH for Tomato Juice

For tomato juice, \([\text{H}^+] = 3.2 \times 10^{-4}\) M.
Using the pH formula:
\( \text{pH} = -\log_{10}(3.2 \times 10^{-4}) \)
Calculate \( \log_{10}(3.2 \times 10^{-4}) = \log_{10}(3.2) + \log_{10}(10^{-4}) \)
\( \log_{10}(3.2) \approx 0.5051 \) and \( \log_{10}(10^{-4}) = -4 \)
So, \( \log_{10}(3.2 \times 10^{-4}) \approx 0.5051 - 4 = -3.4949 \)
Then, \( \text{pH} = -(-3.4949) \approx 3.5 \) (rounded to one decimal place)

Step4: Calculate pH for Seawater

For seawater, \([\text{H}^+] = 5.0 \times 10^{-9}\) M.
Using the pH formula:
\( \text{pH} = -\log_{10}(5.0 \times 10^{-9}) \)
Calculate \( \log_{10}(5.0 \times 10^{-9}) = \log_{10}(5.0) + \log_{10}(10^{-9}) \)
\( \log_{10}(5.0) \approx 0.6990 \) and \( \log_{10}(10^{-9}) = -9 \)
So, \( \log_{10}(5.0 \times 10^{-9}) \approx 0.6990 - 9 = -8.3010 \)
Then, \( \text{pH} = -(-8.3010) \approx 8.3 \) (rounded to one decimal place)

Answer:

(a) The pH of oranges is approximately 3.6.
(b) The pH of tomato juice is approximately 3.5.
(c) The pH of seawater is approximately 8.3.