SME
Principles of Biology I · Problem type
pH and how many times more acidic
Use this for “How many times more acidic is pH 2 than pH 7.4?”
Diagram
Since pH = −log₁₀[H⁺], here [H⁺] = 10⁻⁷·⁴⁰ = 3.9811 × 10⁻⁸ mol/L.
Above 7: basic. pOH = 14 − pH = 6.6.
Second pH pH₂ = 2.00
Times more acidic r = 251,189
This is an example. Type your own values to start a new problem.
r
=
10
−
[H⁺]₁
First [H⁺]
calculated · M
Formulas
pH₁ = −log₁₀([H⁺]₁)
7.40
=
−log₁₀(3.9811
×
10
-8
)
r
=
10
pH₁ − pH₂
251,189
=
10
7.40 − 2.00
Assumptions
•
pH = −log₁₀[H⁺] with [H⁺] in mol/L, so each pH unit is a tenfold change in [H⁺].
•
r = [H⁺]₂ ÷ [H⁺]₁ is how many times more acidic the second solution is; below 1 it is less acidic.
•
Below pH 7 a solution is acidic, above 7 basic (water at 25 °C).
Step-by-step
WE KNOW
pH₁ = 7.40, pH₂ = 2.00
FIND
first [H⁺] ([H⁺]₁), times more acidic (r)
Step 1 · Find first [H⁺] ([H⁺]₁)
Use pH₁ = −log₁₀([H⁺]₁)
Undo the log: one over 10 to the power of the pH.
[H⁺]₁ = 1 ÷ 10pH₁
[H⁺]₁
=
1
10
7.40
[H⁺]₁
=
1
25,120,000
[H⁺]₁
=
3.9811 ×
10
-8
M
Step 2 · Find times more acidic (r)
Use r = 10pH₁ − pH₂
Each pH unit lower is 10 times the [H⁺], so raise 10 to the pH difference.
r
=
10
7.40 − 2.00
r
=
10
5.4
r
=
251,189
Check
Put the numbers back in. Both sides should match.
7.40
=
−log₁₀(3.9811
×
10
-8
)
✓
251,189
=
10
7.40 − 2.00
✓
Topic check
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