The concentration of H+ in a glass of milk is around 3.2 x 10-7 The pH must then be around (Round to the 2nd decimal)

Chemistry: Principles and Reactions
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Chapter13: Acids And Bases
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Problem 76QAP: A solution of sodium cyanide, NaCN, has a pH of 12.10. How many grams of NaCN are in 425 mL of a...
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**Calculating the pH of Milk**

To understand the acidity or alkalinity of a solution, we often use the pH scale. The concentration of hydrogen ions (H+) plays a crucial role in determining the pH value of a solution.

In this example, we are provided with the concentration of H+ ions in a glass of milk, which is approximately \(3.2 \times 10^{-7}\).

The pH of a solution is calculated using the formula:
\[ \text{pH} = -\log[H^+] \]

Given:
\[ [H^+] = 3.2 \times 10^{-7} \]

Substituting the given concentration into the pH formula:

\[ \text{pH} = -\log(3.2 \times 10^{-7}) \]

**Step-by-Step Calculation:**

1. Calculate the logarithm of the concentration:
   \[ \log(3.2 \times 10^{-7}) = \log(3.2) + \log(10^{-7}) \]
   
2. Use the properties of logarithms:
   \[ \log(3.2 \times 10^{-7}) = \log(3.2) + (-7) \]
   
3. Approximate the logarithm of 3.2 (usually, \(\log(3.2) \approx 0.51\)):
   \[ \log(3.2 \times 10^{-7}) = 0.51 - 7 \]
   
4. Combine the results:
   \[ 0.51 - 7 = -6.49 \]

5. The pH is the negative of this value:
   \[ \text{pH} = -(-6.49) = 6.49 \]
   
Thus, the pH of the milk would be around **6.49** (rounded to the second decimal).
Transcribed Image Text:**Calculating the pH of Milk** To understand the acidity or alkalinity of a solution, we often use the pH scale. The concentration of hydrogen ions (H+) plays a crucial role in determining the pH value of a solution. In this example, we are provided with the concentration of H+ ions in a glass of milk, which is approximately \(3.2 \times 10^{-7}\). The pH of a solution is calculated using the formula: \[ \text{pH} = -\log[H^+] \] Given: \[ [H^+] = 3.2 \times 10^{-7} \] Substituting the given concentration into the pH formula: \[ \text{pH} = -\log(3.2 \times 10^{-7}) \] **Step-by-Step Calculation:** 1. Calculate the logarithm of the concentration: \[ \log(3.2 \times 10^{-7}) = \log(3.2) + \log(10^{-7}) \] 2. Use the properties of logarithms: \[ \log(3.2 \times 10^{-7}) = \log(3.2) + (-7) \] 3. Approximate the logarithm of 3.2 (usually, \(\log(3.2) \approx 0.51\)): \[ \log(3.2 \times 10^{-7}) = 0.51 - 7 \] 4. Combine the results: \[ 0.51 - 7 = -6.49 \] 5. The pH is the negative of this value: \[ \text{pH} = -(-6.49) = 6.49 \] Thus, the pH of the milk would be around **6.49** (rounded to the second decimal).
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