Question

for each of the noted items on the ph scale, match the label that indicates (roughly) how many times more acidic they are compared to water.
labels: (10^+), (100^+), (1,000^+), (10,000^+)
drag each item above to its appropriate location in the image. note that every item may not have a match, while some items may have more than one match.
ph scale (vertical):

• 8: seawater (7.5–8.3)
• human blood (7.4)
• 7 (neutral): pure water
• milk (6.5)
• 6
• natural rainwater (5.6)
• 5
• tomatoes (4.5)
• 4
• oranges (3.5)
• 3
• lemons (2.3)
• 2: stomach acid (1.5–2.0)
• 1

Explanation:

The pH scale is logarithmic, meaning each whole number change in pH represents a tenfold change in acidity (or basicity). Pure water has a pH of 7. To determine how many times more acidic a substance is than water, we calculate the difference in pH from 7 and use \(10^{\text{pH difference}}\).

Step 1: Understand the pH difference formula

For a substance with pH \(p\), the acidity factor relative to water (pH 7) is \(10^{7 - p}\) (since lower pH means more acidic, so we subtract the substance's pH from 7 to get the exponent).

Step 2: Analyze each substance:

- **Milk (pH 6.5):**
pH difference: \(7 - 6.5 = 0.5\)? Wait, no—wait, the labels are \(10^+\), \(100^+\), \(1000^+\), \(10,000^+\), which correspond to \(10^1\), \(10^2\), \(10^3\), \(10^4\) (since each pH unit is a factor of 10). Wait, actually, the pH scale: each decrease of 1 pH unit is 10x more acidic. So:
- Water: pH 7.
- Milk: pH 6.5 → difference from 7 is \(7 - 6.5 = 0.5\)? No, the labels are \(10^+\) (10x), \(100^+\) (100x = \(10^2\)), \(1000^+\) (\(10^3\)), \(10,000^+\) (\(10^4\)). Let's list substances by pH and calculate \(10^{7 - \text{pH}}\):

- **Milk (6.5):** \(7 - 6.5 = 0.5\) → \(10^{0.5} \approx 3.16\) (not matching labels, maybe the problem uses whole pH units for simplicity).
- **Natural rainwater (5.6):** \(7 - 5.6 = 1.4\) → ~\(10^{1.4} \approx 25\) (no). Wait, maybe the problem rounds to whole pH differences. Let's check whole numbers:

- **Milk (6.5) ≈ pH 7? No, 6.5 is 0.5 below 7. Maybe the problem uses the closest whole number difference. Wait, the labels are \(10^+\) (10x), \(100^+\) (100x), \(1000^+\) (1000x), \(10,000^+\) (10,000x), which are \(10^1\), \(10^2\), \(10^3\), \(10^4\). So:

- **Milk (pH 6.5):** pH difference from 7 is \(7 - 6.5 = 0.5\) → not a whole number. Maybe the problem considers pH 6 (difference 1: \(10^1 = 10^+\)), pH 5 (difference 2: \(10^2 = 100^+\)), pH 4 (difference 3: \(10^3 = 1000^+\)), pH 3 (difference 4: \(10^4 = 10,000^+\))? Wait, no—wait, lower pH is more acidic. So:

- **Milk (pH 6.5):** closer to pH 7, but maybe the problem groups by pH values:

Let's list substances with their pH and the factor:

- **Milk (6.5):** pH 6.5 → difference from 7 is 0.5 → ~10^0.5 ≈ 3 (not matching). Maybe the problem uses the pH value's integer part:

- **Milk (6.5):** pH 6 (integer) → difference 1 → 10^1 = 10^+
- **Natural rainwater (5.6):** pH 5 (integer) → difference 2 → 10^2 = 100^+
- **Tomatoes (4.5):** pH 4 (integer) → difference 3 → 10^3 = 1000^+
- **Oranges (3.5):** pH 3 (integer) → difference 4 → 10^4 = 10,000^+
- **Lemons (2.3):** pH 2 (integer) → difference 5 → 10^5 (not a label)
- **Stomach acid (1.5–2.0):** pH 2 (integer) → difference 5 (not a label)
- **Seawater (7.5–8.3):** pH 8 (integer) → difference -1 (more basic, so not acidic)
- **Human blood (7.4):** pH 7 (neutral) → difference 0 (not acidic)

Wait, maybe the problem is structured as: each time the pH decreases by 1 (from 7), acidity increases by 10x. So:

- pH 6: 10x more acidic than water (\(10^+\))
- pH 5: 100x (\(100^+\))
- pH 4: 1000x (\(1000^+\))
- pH 3: 10,000x (\(10,000^+\))

Now match substances to their pH (rounded to nearest integer or by their pH value):

- **Milk (6.5):** pH ~6 → 10x (\(10^+\))
- **Natural rainwater (5.6):** pH ~5 → 100x (\(100^+\))
- **Tomatoes (4.5):** pH ~4 → 1000x (\(1000^+\))
- **Oranges (3.5):** pH ~3 → 10,000x (\(10,000^+\))

So the matches are:

- \(10^+\): Milk (6.5), maybe Natural rainwater? No, 5.6 is closer to 5. Wait, maybe the problem uses the pH value's distance from 7:

- Milk (6.5): 7 - 6.5 = 0.5 → 10^0.5 ≈ 3 (no). Wait, the labels are \(10^+\), \(100^+\), \(1000^+\), \(10,000^+\), which are 10^1, 10^2, 10^3, 10^4. So the exponent is (7 - pH). Let's calculate (7 - pH) for each:

- Milk (6.5): 7 - 6.5 = 0.5 → 10^0.5 ≈ 3 (not a label)
- Natural rainwater (5.6): 7 - 5.6 = 1.4 → 10^1.4 ≈ 25 (no)
- Tomatoes (4.5): 7 - 4.5 = 2.5 → 10^2.5 ≈ 316 (no)
- Oranges (3.5): 7 - 3.5 = 3.5 → 10^3.5 ≈ 3162 (no)
- Lemons (2.3): 7 - 2.3 = 4.7 → 10^4.7 ≈ 50,000 (no)
- Stomach acid (1.5–2.0): 7 - 1.75 ≈ 5.25 → 10^5.25 ≈ 178,000 (no)

Wait, maybe the problem is simplified: each pH unit below 7 is a factor of 10. So:

- pH 6: 10x (10^+)
- pH 5: 100x (100^+)
- pH 4: 1000x (1000^+)
- pH 3: 10,000x (10,000^+)

So:

- Milk (6.5) is between pH 6 and 7 → 10x (10^+)
- Natural rainwater (5.6) is between pH 5 and 6 → 100x (100^+)
- Tomatoes (4.5) is between pH 4 and 5 → 1000x (1000^+)
- Oranges (3.5) is between pH 3 and 4 → 10,000x (10,000^+)

Thus, the matches are:

- \(10^+\): Milk (6.5)
- \(100^+\): Natural rainwater (5.6)
- \(1000^+\): Tomatoes (4.5)
- \(10,000^+\): Oranges (3.5)

(Note: The problem says "some items may have more than one match" or "every item may not have a match", but this is the logical breakdown using the logarithmic pH scale.)

Answer:

• \(10^+\) → Milk (6.5)
• \(100^+\) → Natural rainwater (5.6)
• \(1000^+\) → Tomatoes (4.5)
• \(10,000^+\) → Oranges (3.5)

(Other substances like lemons, stomach acid, seawater, blood have pH values that don’t align with the given labels, as their acidity factors exceed or don’t match \(10^+\), \(100^+\), \(1000^+\), \(10,000^+\).)