pH Calculations worksheet with completed examples showing acid and base identification.

The image you've uploaded is a pH calculation worksheet from "The Clever Chemist" (2018). The task involves using the relationship between pH, pOH, and ion concentrations ([H⁺] and [OH⁻]) to complete a table. The key formula used is:

\[
\text{pH} + \text{pOH} = 14
\]

Also:
- \(\text{pH} = -\log[\text{H}^+]\)
- \(\text{pOH} = -\log[\text{OH}^-]\)

We can use these formulas to fill in missing values.

Let’s go through each row and verify or calculate the missing values.

---

Row 1:

- \([H^+] = 1 \times 10^{-10}\)
- pH = \( -\log(1 \times 10^{-10}) = 10 \) ✔
- \([OH^-] = 1 \times 10^{-4}\)
- pOH = \( -\log(1 \times 10^{-4}) = 4 \) ✔
- pH + pOH = 10 + 4 = 14 ✔
- Acid or Base? pOH = 4 → pH = 10 → Basic ✔

✔ All correct.

---

Row 2:

- \([H^+] = 2.5 \times 10^{-9}\)
- pH = \( -\log(2.5 \times 10^{-9}) \approx 8.6 \) ✔
- \([OH^-] = 4 \times 10^{-6}\)
- pOH = \( -\log(4 \times 10^{-6}) \approx 5.4 \)
- Check: pH + pOH = 8.6 + 5.4 = 14 ✔
- Acid or Base? pH > 7 → Basic ✔

But the checkmark is red, indicating it's marked wrong — but calculations are correct. Maybe a typo?

Wait — let’s double-check:
- \([H^+] = 2.5 \times 10^{-9}\), so \(-\log(2.5 \times 10^{-9}) = 9 - \log(2.5) = 9 - 0.3979 ≈ 8.602\) → ≈8.6 ✔
- \([OH^-] = 4 \times 10^{-6}\), so pOH = \( -\log(4 \times 10^{-6}) = 6 - \log(4) = 6 - 0.602 ≈ 5.398\) → ≈5.4 ✔
- Sum = 8.6 + 5.4 = 14 ✔

So why red? Maybe inconsistency in significant figures? Or perhaps the expected answer was different?

But it seems correct. Let’s keep going.

---

Row 3:

- \([H^+] = 1 \times 10^{-1}\) → pH = \( -\log(0.1) = 1 \) ✔
- \([OH^-] = 1 \times 10^{-13}\) → pOH = 13 ✔
- pH + pOH = 1 + 13 = 14 ✔
- pH = 1 → Acidic ✔

Red checkmark — but correct.

---

Row 4:

- \([H^+] = 2 \times 10^{-4}\) → pH = \( -\log(2 \times 10^{-4}) = 4 - \log(2) ≈ 4 - 0.3 = 3.7 \) ✔
- \([OH^-] = 2 \times 10^{-10}\) → pOH = \( -\log(2 \times 10^{-10}) = 10 - \log(2) ≈ 9.7 \) → Wait! But written as 10.3

Wait! This is incorrect.

Given \([OH^-] = 2 \times 10^{-10}\), then:
- pOH = \( -\log(2 \times 10^{-10}) = 10 - \log(2) ≈ 10 - 0.3010 = 9.699 ≈ 9.7 \)

But the table says 10.3 → ✘ Incorrect

Then pH = 14 - pOH = 14 - 9.7 = 4.3 → But table says 3.7 → So both pH and pOH are wrong.

Wait, but pH is given as 3.7, which corresponds to \([H^+] = 2 \times 10^{-4}\), so that’s correct.

But if \([H^+] = 2 \times 10^{-4}\), then \([OH^-] = \frac{K_w}{[H^+]} = \frac{1 \times 10^{-14}}{2 \times 10^{-4}} = 5 \times 10^{-11}\)

But the table says \([OH^-] = 2 \times 10^{-10}\) → That’s wrong.

So:
- Correct \([OH^-] = 5 \times 10^{-11}\)
- Then pOH = \( -\log(5 \times 10^{-11}) = 11 - \log(5) ≈ 11 - 0.699 = 10.301 ≈ 10.3 \)

Ah! So the table has:
- \([OH^-] = 2 \times 10^{-10}\) → incorrect
- But pOH = 10.3 → which would be correct for \([OH^-] = 5 \times 10^{-11}\)

So there’s an inconsistency.

Let’s recalculate properly:

Given:
- \([H^+] = 2 \times 10^{-4}\)
- pH = \( -\log(2 \times 10^{-4}) = 3.7 \) ✔
- \([OH^-] = \frac{1 \times 10^{-14}}{2 \times 10^{-4}} = 5 \times 10^{-11}\) → should be this
- pOH = \( -\log(5 \times 10^{-11}) = 10.3 \) ✔
- pH + pOH = 3.7 + 10.3 = 14 ✔

But table says \([OH^-] = 2 \times 10^{-10}\) → that’s wrong — it should be \(5 \times 10^{-11}\)

So the error is in \([OH^-]\) column.

Similarly, next row...

---

Row 5:

- \([H^+] = 2 \times 10^{-1}\) → pH = \( -\log(0.2) = 0.7 \) → but table says 6.7 ✘

Wait! That’s clearly wrong.

\([H^+] = 2 \times 10^{-1} = 0.2 M\), so pH = \( -\log(0.2) = 0.7 \), not 6.7.

But table says pH = 6.7 → Wrong

Then pOH = 14 - 6.7 = 7.3 → but \([OH^-] = 5 \times 10^{-7}\) → pOH = \( -\log(5 \times 10^{-7}) = 6.3 \) → not 7.3

Inconsistency.

Let’s fix:

- \([H^+] = 2 \times 10^{-1} = 0.2\)
- pH = \( -\log(0.2) = 0.7 \)
- \([OH^-] = \frac{1 \times 10^{-14}}{0.2} = 5 \times 10^{-14}\)
- pOH = \( -\log(5 \times 10^{-14}) = 13.3 \)
- pH + pOH = 0.7 + 13.3 = 14 ✔

So table has:
- pH = 6.7 ✘
- \([OH^-] = 5 \times 10^{-7}\) ✘ → should be \(5 \times 10^{-14}\)
- pOH = 7.3 ✘ → should be 13.3

So multiple errors here.

But wait — maybe the row is misread.

Look at the table:

Row 5:
- \([H^+] = 2 \times 10^{-1}\)
- pH = 6.7 ✘
- \([OH^-] = 5 \times 10^{-7}\)
- pOH = 7.3
- pH + pOH = 14 → 6.7 + 7.3 = 14 ✔

But \([H^+] = 2 \times 10^{-1}\) → pH = 0.7, not 6.7

And if \([OH^-] = 5 \times 10^{-7}\), pOH = 6.3, not 7.3

So inconsistency.

But 6.7 + 7.3 = 14 → math checks out, but values don’t match.

So either:
- \([H^+]\) is wrong
- or pH is wrong

But \([H^+] = 2 \times 10^{-1}\) → pH = 0.7

Unless it's a typo in the table.

Wait — perhaps the \([H^+]\) is actually \(2 \times 10^{-7}\)? Because:
- If \([H^+] = 2 \times 10^{-7}\), pH = \( -\log(2 \times 10^{-7}) = 6.7 \)
- Then \([OH^-] = 10^{-14}/(2 \times 10^{-7}) = 5 \times 10^{-8}\)
- pOH = \( -\log(5 \times 10^{-8}) = 7.3 \)
- pH + pOH = 6.7 + 7.3 = 14 ✔

Ah! So likely, the \([H^+]\) value is mistyped — it should be \(2 \times 10^{-7}\), not \(2 \times 10^{-1}\)

So the error is in the \([H^+]\) column.

Similarly, check Row 6:

- \([H^+] = 6.3 \times 10^{-10}\)
- pH = \( -\log(6.3 \times 10^{-10}) = 9.2 \) → but table says 13.2 ✘

Wait — 6.3 × 10⁻¹⁰ → log(6.3) ≈ 0.8, so pH = 10 - 0.8 = 9.2

But table says pH = 13.2 → too high

Then \([OH^-] = 1.6 \times 10^{-5}\) → pOH = \( -\log(1.6 \times 10^{-5}) = 4.8 \)

But 13.2 + 4.8 = 18 → not 14 → invalid

So something is wrong.

Wait — let’s compute:

If \([H^+] = 6.3 \times 10^{-10}\), then:
- pH = \( -\log(6.3 \times 10^{-10}) = 9.2 \)
- \([OH^-] = 10^{-14}/(6.3 \times 10^{-10}) = 1.587 \times 10^{-5} ≈ 1.6 \times 10^{-5}\)
- pOH = \( -\log(1.6 \times 10^{-5}) = 4.8 \)
- pH + pOH = 9.2 + 4.8 = 14 ✔

So table has:
- pH = 13.2 ✘ → should be 9.2
- pOH = 0.8 ✘ → should be 4.8

But 13.2 + 0.8 = 14 → sum is correct, but values are off.

So again, likely a typo in pH or pOH.

Wait — perhaps \([H^+]\) is \(6.3 \times 10^{-14}\)? Then:
- pH = \( -\log(6.3 \times 10^{-14}) = 13.2 \)
- \([OH^-] = 10^{-14}/(6.3 \times 10^{-14}) = 0.1587 ≈ 1.6 \times 10^{-1}\)
- pOH = \( -\log(1.6 \times 10^{-1}) = 0.8 \)
- pH + pOH = 13.2 + 0.8 = 14 ✔

So yes — the \([H^+]\) must be \(6.3 \times 10^{-14}\), not \(6.3 \times 10^{-10}\)

So typo in \([H^+]\) column.

Similarly, Row 7:
- \([H^+] = 3.2 \times 10^{-10}\)
- pH = \( -\log(3.2 \times 10^{-10}) = 9.5 \)
- Table says pH = 2.5 ✘

But if pH = 2.5, then \([H^+] = 10^{-2.5} = 3.16 \times 10^{-3}\), not 3.2 × 10⁻¹⁰

So again, mismatch.

But table says:
- \([H^+] = 3.2 \times 10^{-10}\)
- pH = 2.5 ✘
- \([OH^-] = 3.2 \times 10^{-5}\)
- pOH = 11.5 → \( -\log(3.2 \times 10^{-5}) = 4.5 \) → not 11.5

Wait — pOH = 11.5 → \([OH^-] = 10^{-11.5} = 3.16 \times 10^{-12}\)

But table says \([OH^-] = 3.2 \times 10^{-5}\)

So no consistency.

But 2.5 + 11.5 = 14 → sum is correct.

So likely:
- pH = 2.5 → \([H^+] = 3.2 \times 10^{-3}\)
- pOH = 11.5 → \([OH^-] = 3.2 \times 10^{-12}\)

But table says \([H^+] = 3.2 \times 10^{-10}\), which is wrong.

So again, typo in \([H^+]\)

Let’s look for rows with green checkmarks — they’re correct.

Row 8:
- \([H^+] = 1 \times 10^{-9}\)
- pH = 9 ✔
- \([OH^-] = 1 \times 10^{-5}\)
- pOH = 5 ✔
- pH + pOH = 14 ✔
- Basic ✔

Check: \([H^+] = 10^{-9}\), so \([OH^-] = 10^{-14}/10^{-9} = 10^{-5}\) → correct ✔

Row 9:
- \([H^+] = 1.6 \times 10^{-1}\)
- pH = \( -\log(0.16) ≈ 0.8 \) → table says 0.8 ✔
- \([OH^-] = 6.3 \times 10^{-14}\)
- pOH = \( -\log(6.3 \times 10^{-14}) = 13.2 \)
- pH + pOH = 0.8 + 13.2 = 14 ✔
- Acidic ✔

Correct.

Row 10:
- \([H^+] = 5 \times 10^{-1}\)
- pH = \( -\log(0.5) = 0.3 \) → table says 4.3 ✘

Wait — 0.5 → log(0.5) = -0.301 → pH = 0.3

But table says 4.3 → wrong

But \([OH^-] = 2 \times 10^{-14}\) → pOH = 13.7 → pH = 14 - 13.7 = 0.3 → so pH should be 0.3

But table says 4.3 → error

So likely, \([H^+]\) is \(5 \times 10^{-5}\)? Then:
- pH = \( -\log(5 \times 10^{-5}) = 4.3 \)
- \([OH^-] = 10^{-14}/(5 \times 10^{-5}) = 2 \times 10^{-10}\)
- pOH = \( -\log(2 \times 10^{-10}) = 9.7 \)
- pH + pOH = 4.3 + 9.7 = 14 ✔

So \([OH^-]\) should be \(2 \times 10^{-10}\), not \(2 \times 10^{-14}\)

So typo in \([OH^-]\)

Row 11:
- \([H^+] = 1.3 \times 10^{-13}\)
- pH = \( -\log(1.3 \times 10^{-13}) = 12.9 \) → table says 12.9 ✔
- \([OH^-] = 7.9 \times 10^{-2}\)
- pOH = \( -\log(7.9 \times 10^{-2}) = 1.1 \)
- pH + pOH = 12.9 + 1.1 = 14 ✔
- Basic ✔

Check: \([H^+] = 1.3 \times 10^{-13}\), so \([OH^-] = 10^{-14}/(1.3 \times 10^{-13}) = 0.0769 ≈ 7.7 \times 10^{-2}\) → close to 7.9 × 10⁻² → rounding acceptable

So correct.

Row 12:
- \([H^+] = 4 \times 10^{-8}\)
- pH = \( -\log(4 \times 10^{-8}) = 7.4 \) ✔
- \([OH^-] = 2.5 \times 10^{-7}\)
- pOH = \( -\log(2.5 \times 10^{-7}) = 6.6 \) ✔
- pH + pOH = 7.4 + 6.6 = 14 ✔
- Neutral? pH = 7.4 → slightly basic → but table says basic ✔

Correct.

---

Summary of Errors:

The red checkmarks indicate errors in the table, but many of them are due to typos in the input values.

Here are the corrected values for rows with red marks:

#### Row 4:
- \([H^+] = 2 \times 10^{-4}\) → pH = 3.7 ✔
- \([OH^-] = 5 \times 10^{-11}\) (not 2 × 10⁻¹⁰)
- pOH = 10.3 ✔
- Acidic ✔

#### Row 5:
- \([H^+] = 2 \times 10^{-7}\) (not 2 × 10⁻¹)
- pH = 6.7 ✔
- \([OH^-] = 5 \times 10^{-8}\) (not 5 × 10⁻⁷)
- pOH = 7.3 ✔
- Basic ✔

#### Row 6:
- \([H^+] = 6.3 \times 10^{-14}\) (not 6.3 × 10⁻¹⁰)
- pH = 13.2 ✔
- \([OH^-] = 1.6 \times 10^{-1}\) (0.16 M)
- pOH = 0.8 ✔
- Basic ✔

#### Row 7:
- \([H^+] = 3.2 \times 10^{-3}\) (not 3.2 × 10⁻¹⁰)
- pH = 2.5 ✔
- \([OH^-] = 3.2 \times 10^{-12}\)
- pOH = 11.5 ✔
- Acidic ✔

#### Row 10:
- \([H^+] = 5 \times 10^{-5}\) (not 5 × 10⁻¹)
- pH = 4.3 ✔
- \([OH^-] = 2 \times 10^{-10}\) (not 2 × 10⁻¹⁴)
- pOH = 9.7 → but table says 9.7? No, it says pOH = 9.7? Wait, table says pOH = 9.7?

Wait — Row 10: pOH = 9.7 → but table says 9.7? No, table says pOH = 9.7? Let me check.

Table says:
- \([H^+] = 5 \times 10^{-1}\)
- pH = 4.3
- \([OH^-] = 2 \times 10^{-14}\)
- pOH = 9.7
- pH + pOH = 14 → 4.3 + 9.7 = 14 ✔

But \([OH^-] = 2 \times 10^{-14}\) → pOH = \( -\log(2 \times 10^{-14}) = 13.7 \), not 9.7

So pOH should be 13.7, not 9.7

But 4.3 + 9.7 = 14 → sum is correct, but values inconsistent.

So likely:
- pH = 4.3 → \([H^+] = 5 \times 10^{-5}\)
- pOH = 9.7 → \([OH^-] = 2 \times 10^{-10}\)
- Then \([H^+] = 10^{-14}/(2 \times 10^{-10}) = 5 \times 10^{-5}\) → matches

So correct values:
- \([H^+] = 5 \times 10^{-5}\)
- \([OH^-] = 2 \times 10^{-10}\)
- pH = 4.3
- pOH = 9.7

But table says \([H^+] = 5 \times 10^{-1}\) and \([OH^-] = 2 \times 10^{-14}\) → both wrong

So typo in both columns.

---

✔ Final Answer:

The worksheet contains several typos in the \([H^+]\) and \([OH^-]\) columns. The red checkmarks indicate incorrect entries, but the sum pH + pOH = 14 is maintained by adjusting other values incorrectly.

To solve the problem correctly:

Use:
- \([H^+] = 10^{-\text{pH}}\)
- \([OH^-] = 10^{-\text{pOH}}\)
- \([H^+][OH^-] = 1 \times 10^{-14}\)
- pH + pOH = 14

Correct the following rows:

| Row | Correction |
|-----|-----------|
| 4 | \([OH^-] = 5 \times 10^{-11}\) |
| 5 | \([H^+] = 2 \times 10^{-7}\), \([OH^-] = 5 \times 10^{-8}\) |
| 6 | \([H^+] = 6.3 \times 10^{-14}\), \([OH^-] = 1.6 \times 10^{-1}\) |
| 7 | \([H^+] = 3.2 \times 10^{-3}\), \([OH^-] = 3.2 \times 10^{-12}\) |
| 10 | \([H^+] = 5 \times 10^{-5}\), \([OH^-] = 2 \times 10^{-10}\) |

All other rows are correct.

Conclusion: The task is to identify and correct the inconsistencies using the pH-pOH relationship and ion product of water (\(K_w = 1 \times 10^{-14}\)).

The red checkmarks highlight where the given values contradict the rules of acid-base chemistry.