SME
Grade 10 Science · Chemistry
Acids, bases and pH
Use this for “[H⁺] = 2.5 × 10⁻⁴ M. What are the pH, the pOH and [OH⁻]?”
Diagram
Since pH = −log₁₀[H⁺], here [H⁺] = 10⁻³·⁶⁰² = 0.00025 mol/L.
Below 7: acidic.
This is an example. Type your own values to start a new problem.
pH
=
−log₁₀(
[H⁺]
)
pOH
pOH
calculated
[OH⁻]
Hydroxide ion concentration
calculated · mol/L
Formulas
pH = −log₁₀([H⁺])
3.602
=
−log₁₀(2.5
×
10
-4
)
pH + pOH = 14
3.602 + 10.4 = 14
[OH⁻]
=
(1
×
10
-14
)/[H⁺]
4.00 ×
10
-11
=
(1
×
10
-14
)/(2.5
×
10
-4
)
Assumptions
•
The water is at 25 °C, where [H⁺][OH⁻] = 1 × 10⁻¹⁴ and pH + pOH = 14.
•
Each pH unit is ten times the [H⁺]. Below 7 is acidic, above 7 basic.
Step-by-step
WE KNOW
[H⁺] = 2.5 × 10⁻⁴ mol/L
FIND
pH (pH), pOH (pOH), hydroxide ion concentration ([OH⁻])
Step 1 · Find pH (pH)
Use pH = −log₁₀([H⁺])
Each step of 1 on the scale is ten times the [H⁺]: the log counts the powers of ten.
pH
=
−log₁₀(2.5
×
10
-4
)
pH = 3.602
Step 2 · Find pOH (pOH)
Use pH + pOH = 14
In water at 25 °C, pH and pOH add to 14.
pOH = 14 − pH
pOH = 14 − 3.602
pOH = 10.4
Step 3 · Find hydroxide ion concentration ([OH⁻])
Use [OH⁻] = (1 × 10⁻¹⁴)/[H⁺]
In
water
[H⁺]
×
[OH⁻]
is
always
1 ×
10
-14
.
[OH⁻]
=
10
-14
/(2.5
×
10
-4
)
[OH⁻]
=
4.00 ×
10
-11
mol/L
Check
Put the numbers back in. Both sides should match.
3.602
≈
−log₁₀(2.5
×
10
-4
)
✓
3.602 + 10.4 = 14 ✓
4.00 ×
10
-11
=
10
-14
/(2.5
×
10
-4
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Refresh
More problem types
[H⁺] from pH
Use this for “A solution has a pH of 8.50. What is its [H⁺]?”
A strong base: pOH and pH
Use this for “A solution of 0.010 M NaOH: what are its pOH and pH?”
Titrating a strong acid
Use this for “25 mL of 0.1 M HCl is titrated with 0.2 M NaOH. Where is the equivalence point?”
Titrating a weak acid
Use this for “20 mL of 0.1 M acetic acid (Kₐ = 1.8 × 10⁻⁵) is titrated with 0.1 M NaOH.”
Acid, base or neutral?
Use this for “Is baking soda solution an acid, a base or neutral?”