pH Calculations worksheet for chemistry students, featuring problems on determining pH of different solutions.

Let’s solve each problem step by step. We’ll use the definition:
pH = -log[H⁺], where [H⁺] is the concentration of hydrogen ions in moles per liter (M).

For bases, we first find pOH using [OH⁻], then use:
pH + pOH = 14 → so pH = 14 - pOH

Also remember:
- Strong acids like HNO₃, HCl, and H₂SO₄ (first proton) dissociate completely.
- Strong bases like NaOH and Sr(OH)₂ also dissociate completely.
- For H₂SO₄, it gives 2 H⁺ per molecule — but only the first proton is strong; however, for simplicity at this level, many textbooks treat both protons as fully dissociating unless specified otherwise. But let’s check context: since it says “calculate pH” without mentioning Ka2, we’ll assume full dissociation for both protons? Actually — wait — standard practice in intro chem: for dilute solutions of H₂SO₄, we often consider only the first H⁺ as strong, and second is weak. BUT looking at the problems here, they’re likely expecting simple stoichiometry. Let me think...

Actually, let’s look at problem #3: 0.0025 M H₂SO₄. If we assume complete dissociation of both protons, [H⁺] = 2 × 0.0025 = 0.0050 M → pH = -log(0.0050) ≈ 2.30

But if only first proton: [H⁺] = 0.0025 → pH = 2.60

Which one to choose? Since this is a basic worksheet and no Ka values are given, and other problems are straightforward, I think they expect us to treat H₂SO₄ as giving 2 H⁺. However, let’s verify with common textbook approach.

Wait — actually, in many high school or early college courses, when calculating pH of H₂SO₄ without additional info, they may expect you to use [H⁺] = 2 × concentration. But technically, that’s not accurate because HSO₄⁻ is a weak acid. However, for very dilute solutions (<0.01 M), sometimes approximation is used. Given that 0.0025 M is quite dilute, and no Ka provided, perhaps they want us to assume full dissociation? Or maybe just first proton?

Looking at problem #1: 0.010 M HNO₃ → monoprotic strong acid → [H⁺] = 0.010 → pH = 2.00

Problem #2: 2.5e-6 M HCl → same → [H⁺] = 2.5e-6 → pH = 5.60

Problem #4: 0.0010 M NaOH → [OH⁻] = 0.0010 → pOH = 3.00 → pH = 11.00

Problem #5: 0.020 M Sr(OH)₂ → each formula unit gives 2 OH⁻ → [OH⁻] = 0.040 M → pOH = 1.40 → pH = 12.60

Now back to #3: H₂SO₄. In many introductory worksheets, especially if not specifying partial dissociation, they might expect [H⁺] = 2 × 0.0025 = 0.0050 M → pH = -log(0.0050) = 2.3010 ≈ 2.30

But let’s be precise: actually, for H₂SO₄, the first proton is strong, second has Ka2 = 0.012, so for 0.0025 M, we should do equilibrium. But that’s probably beyond scope. Since all other problems are simple, and no mention of Ka, I think for consistency, we'll go with [H⁺] = 2 × concentration for H₂SO₄. Alternatively, maybe the worksheet expects only first proton? Hmm.

Wait — let's see the fill-in-the-blanks at top: it says "a pH value of less than 7 indicates a(n) ___ solution" — acidic, etc. So it's very basic. Probably, for H₂SO₄, they expect [H⁺] = 2 × Molarity.

I recall that in some curricula, for sulfuric acid, they teach to multiply by 2 for diprotic strong acids. Even though strictly speaking it's not fully accurate, for this level, it's acceptable.

So I'll proceed with [H⁺] = 2 × 0.0025 = 0.0050 M for problem 3.

Let’s calculate each:

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Problem 1: 0.010 M HNO₃

HNO₃ is strong acid → [H⁺] = 0.010 M

pH = -log(0.010) = -log(10⁻²) = 2.00

✔ Final Answer for #1: 2.00

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Problem 2: 2.5 × 10⁻⁶ M HCl

HCl is strong acid → [H⁺] = 2.5 × 10⁻⁶ M

pH = -log(2.5 × 10⁻⁶)

Calculate log(2.5 × 10⁻⁶) = log(2.5) + log(10⁻⁶) = 0.39794 - 6 = -5.60206

So pH = -(-5.60206) = 5.60206 ≈ 5.60 (rounded to two decimal places)

✔ Final Answer for #2: 5.60

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Problem 3: 0.0025 M H₂SO₄

Assuming complete dissociation of both protons (as commonly done in intro courses unless specified):

[H⁺] = 2 × 0.0025 = 0.0050 M

pH = -log(0.0050) = -log(5.0 × 10⁻³) = -(log 5.0 + log 10⁻³) = -(0.69897 - 3) = -(-2.30103) = 2.30103 ≈ 2.30

If they meant only first proton, it would be 2.60, but I think 2.30 is expected here.

To confirm: 0.0050 M → log(0.005) = log(5e-3) = -2.3010 → pH=2.3010 → round to 2.30

✔ Final Answer for #3: 2.30

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Problem 4: 0.0010 M NaOH

NaOH is strong base → [OH⁻] = 0.0010 M

pOH = -log(0.0010) = -log(10⁻³) = 3.00

pH = 14.00 - pOH = 14.00 - 3.00 = 11.00

✔ Final Answer for #4: 11.00

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Problem 5: 0.020 M Sr(OH)₂

Sr(OH)₂ is strong base → each mole gives 2 moles OH⁻

[OH⁻] = 2 × 0.020 = 0.040 M

pOH = -log(0.040) = -log(4.0 × 10⁻²) = -(log 4.0 + log 10⁻²) = -(0.60206 - 2) = -(-1.39794) = 1.39794 ≈ 1.40

pH = 14.00 - 1.40 = 12.60

Check: log(0.04) = log(4e-2) = -1.39794 → pOH=1.39794 → pH=12.60206 → rounds to 12.60

✔ Final Answer for #5: 12.60

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Now, let’s double-check calculations:

#1: 0.010 → log= -2 → pH=2.00 ✔️

#2: 2.5e-6 → log(2.5e-6)= log2.5 + log1e-6 = 0.39794 -6 = -5.60206 → neg is 5.60206 → 5.60 ✔️

#3: 0.0050 → log(0.005)= -2.3010 → pH=2.3010 → 2.30 ✔️

#4: 0.0010 → pOH=3.00 → pH=11.00 ✔️

#5: 0.040 → log(0.04)= -1.39794 → pOH=1.39794 → pH=12.60206 → 12.60 ✔️

All good.

Final Answers:

1. 2.00
2. 5.60
3. 2.30
4. 11.00
5. 12.60

Note: For problem 3, if your teacher expects only the first proton of H₂SO₄ to dissociate completely, then [H⁺] = 0.0025 M → pH = -log(0.0025) = 2.60. But based on typical worksheet expectations and lack of Ka data, 2.30 is more likely intended. If unsure, ask your teacher — but for now, we’ll go with 2.30.

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Final Answer:
1. 2.00
2. 5.60
3. 2.30
4. 11.00
5. 12.60