pH / Acid-Base Calculator

Move the slider or enter a concentration to convert between pH, pOH, [H⁺], and [OH⁻], and see where a solution falls on the acid-base scale.

Worked example — pH / Acid-Base Calculator

pH	7.00
pOH	7.00
[OH-]	1.00×10⁻⁷ mol/L

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Three pH units is a factor of a thousand 🖖

Because pH is a logarithmic measure, each single unit represents a tenfold change in hydrogen-ion concentration — a solution of pH 3 is not "twice as acidic" as pH 6, it holds a thousand times the H⁺ — three units, three factors of ten. This is why a shift of just 0.1 pH units in ocean water, or a swing from pH 7.4 to 7.35 in human blood, represents a real and significant change in ion concentration even though the numbers look small. The scale exists because [H⁺] in everyday solutions spans more than 14 orders of magnitude — from about 10⁻¹⁴ mol/L in strong bases to 10⁰ mol/L in strong acids — and a logarithm compresses that enormous range into a single easy-to-read number from 0 to 14.

pH is really an ion count 🖖

At heart, pH just reports how crowded the free hydrogen ions (H⁺) are in a water-based liquid — the more crowded, the more acidic. Pure water splits slightly into equal amounts of H⁺ and OH⁻ (about 10⁻⁷ mol/L of each at room temperature), which is why neutral lands at pH 7. Since pH + pOH always adds to 14, knowing any one of pH, pOH, [H⁺], or [OH⁻] pins down the other three — which is exactly what this tool does.

Neutral isn't always pH 7 🖖

The neat rules that neutral means pH 7 and that pH + pOH = 14 hold only at 25°C. Water self-ionizes more readily when it is hot, so its ion product K_w grows with temperature: pure water is neutral at about pH 6.8 at body temperature (37°C) and near pH 6 close to boiling. The water is still perfectly neutral — equal H⁺ and OH⁻ — so the famous 7 is really just a room-temperature coincidence.

Problems solved in full

Pure water at pH 7 and a small pH change 5 steps
Pure water is pH 7. Show why, and then work out what a "small" pH change actually means.
1. Water self-ionises, and at 25 °C the product of the two ion concentrations is fixed. In pure water the two are equal by symmetry, so each is the square root of 10⁻¹⁴.
  
  $K_{w} = [H^{+}] [O H^{-}] = 1.0 \times 1 0^{- 14} at 2 5^{\circ} C$
2. pH is minus the base-ten logarithm of that concentration. The logarithm is the whole design: concentrations here span fourteen orders of magnitude, and a linear scale would be unusable.
  
  $pH = - lo g_{10} [H^{+}] = 7.00 \Rightarrow [H^{+}] = 1.0 \times 1 0^{- 7} mol/L$
3. Since the product is fixed, the two p-values must sum to 14. The calculator above prints both, and neutrality means equal — not "7" as such. At 50 °C neutral water is pH 6.63, because K_w is temperature-dependent.
  
  $pH + pOH = 14 \Rightarrow pOH = 7.00, [O H^{-}] = 1.00 \times 1 0^{- 7} mol/L$
4. Now read the scale properly. A drop of three pH units is not a 43% change; each unit is a factor of ten.
  
  $\frac{[ H ^{+} ] _{pH 4}}{[ H ^{+} ] _{pH 7}} = \frac{1 0 ^{- 4}}{1 0 ^{- 7}} = 1000$
5. Apply that to something with a narrow tolerance. Human blood is held between 7.35 and 7.45.
  
  $blood 7.4 \to 7.0 : \frac{1 0 ^{- 7.0}}{1 0 ^{- 7.4}} = 1 0^{0.4} = 2.5 \times$
Answer

pH 7.00, pOH 7.00, both ions at 1.00 × 10⁻⁷ mol/L. pH 7 to pH 4 is a thousandfold increase in acidity, which is why "slightly acidic rain" is not a slight thing. And blood falling from 7.4 to 7.0 — a change that looks like 5% — is a 2.5-fold rise in hydrogen-ion concentration and is fatal. Every intuition about "how much" fails on this scale unless you undo the logarithm first.

$pH = - lo g_{10} [H^{+}] Δ pH = 1 ⟺ 10 \times$
Diluting stomach acid down to the pH of coffee 7 steps
A 250 mL glass of stomach acid sits at pH 1.5. How much water do you have to add to make it as mild as coffee, pH 5.0 — and how far can dilution take you?

Open this problem in the tool
1. Dilution acts on concentration, not on pH, so undo the logarithm at both ends before doing anything else. Stomach acid is hydrochloric, which gives up its proton completely, so the concentration is the whole story.
  
  $[H^{+}]_{1} = 1 0^{- 1.5} = 3.16 \times 1 0^{- 2} mol/L, [H^{+}]_{2} = 1 0^{- 5.00} = 1.00 \times 1 0^{- 5} mol/L$
2. Adding water changes the volume and leaves the amount of acid alone. That is one equation, and it is the only chemistry in the problem.
  
  $c_{1} V_{1} = c_{2} V_{2} ⟹ \frac{V _{2}}{V _{1}} = \frac{c _{1}}{c _{2}} = \frac{1 0 ^{- 1.5}}{1 0 ^{- 5}} = 1 0^{3.5} = 3162$
3. That gap costs a factor of 3162 in volume. Our 250 mL glass has to become 791 litres, which is roughly four bathtubs of water for one glass of acid.
  
  $V_{2} = 0.250 L \times 3162 = 791 L$
4. Set the calculator to pH 5.00 and check the rest of the picture: pOH 9.00, and the hydroxide concentration four orders of magnitude below the hydrogen-ion concentration in the same glass. Hydroxide has in fact climbed on the way here, by the same factor the acid fell.
  
  $pOH = 14 - 5.00 = 9.00, [O H^{-}] = 1.00 \times 1 0^{- 9} mol/L$
5. Now push further, because the rule is easy to over-trust. Coffee to neutral is only two more pH units, and two more units is another factor of 100 — 79,000 litres, which is a backyard swimming pool, for one glass.
  
  $pH 5 \to 7 : \times 1 0^{2} ⟹ 79000 L$
6. And you would still not get there. Every calculation above ignored the hydrogen ions water supplies on its own, which is fine while the acid provides far more. Once the acid is down at water’s own 10⁻⁷ mol/L that assumption is spent, and both sources have to be counted: charge balance gives a quadratic.
  
  $[H^{+}]^{2} - c_{a} [H^{+}] - K_{w} = 0 ⟹ [H^{+}] = \frac{c _{a} + c _{a}^{2} + 4 K _{w}}{2}$
7. Dilute a strong acid until the acid alone would account for pH 7.00, and the solution is actually at pH 6.79. The hydrogen-ion concentration lands on 1.618 × 10⁻⁷ mol/L, which is the golden ratio, for the same algebraic reason it turns up anywhere else: it is the positive root of x² = x + 1 in disguise.
  
  $c_{a} = 1.00 \times 1 0^{- 7} : [H^{+}] = 1.618 \times 1 0^{- 7} mol/L \Rightarrow pH = 6.79$
Answer

791 litres, and pH 7 is unreachable. Each pH unit is a factor of ten in water, so the cost of mildness grows tenfold per step. Dilution can only carry a strong acid toward neutrality, never through it: at pH 6.79 the water is already contributing six hydrogen ions for every ten the acid contributes, and every litre you add brings more of both.

$V_{2} = V_{1} \cdot 1 0^{Δ pH} = 0.250 \times 1 0^{3.5} = 791 L$

References (1)

The logarithmic definition and why one unit is a factor of ten: D. C. Harris, Quantitative Chemical Analysis, 9th ed., ch. 8. W. H. Freeman, 2015. ISBN 978-1-4641-3538-5.