Worked examples — Ionic product of water Kw = 10⁻¹⁴ at 25 °C

Before we start, one reminder of the two symbols we lean on:

Why ${\log }_{10}$ and not the concentration directly? Because concentrations here span from $10^{-1}$ down to $10^{-14}$ — fourteen powers of ten. Writing "the exponent" (that is exactly what ${\log }_{10}$ hands you, with a minus sign) turns a huge range into friendly numbers between 0 and 14. That is the only reason the pH scale exists.

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The scenario matrix

Every Kw problem is one of these cells. Each row is a different kind of trap.

#	Cell (the scenario)	Sign / regime	What is tricky	Example
A	Acidic: $[{\text{H}}^{+}]>[{\text{OH}}^{-}]$	$[{\text{H}}^{+}]>10^{-7}$	find the small $[{\text{OH}}^{-}]$	Ex 1
B	Basic: $[{\text{OH}}^{-}]>[{\text{H}}^{+}]$	$[{\text{OH}}^{-}]>10^{-7}$	find the small $[{\text{H}}^{+}]$	Ex 2
C	Exactly neutral at 25 °C	$[{\text{H}}^{+}]=[{\text{OH}}^{-}]$	square-root, why they're equal	Ex 3
D	Non-25 °C neutral	$K_w\ne 10^{-14}$	"neutral" is not pH 7	Ex 4
E	Degenerate / limiting: super-dilute acid	$[{\text{H}}^{+}{]}_{\text{added}}\to 10^{-7}$	water's own ions can't be ignored	Ex 5
E'	Degenerate / limiting: super-dilute base	$[{\text{OH}}^{-}{]}_{\text{added}}\to 10^{-7}$	mirror of E, water's H⁺ matters	Ex 5b
F	pH ↔ pOH conversion	any	using $\text{pH}+\text{pOH}=14$	Ex 6
G	Real-world word problem	acidic	translate words → numbers	Ex 7
H	Exam twist: given pOH, non-25 °C	$K_w\ne 10^{-14}$	$\text{pH}+\text{pOH}=\text{p}{K}_w\ne 14$	Ex 8

The seesaw picture below is the mental model behind every cell.

How to read the seesaw. The horizontal beam is balanced on a yellow fulcrum. The left pan (coral) holds $[{\text{H}}^{+}]$; the right pan (lavender) holds $[{\text{OH}}^{-}]$. The rule printed across the top, $[{\text{H}}^{+}][{\text{OH}}^{-}]=K_w=10^{-14}$, is the invisible constraint that keeps the beam obeying one law: whenever you push the coral H⁺ pan up (adding acid, the coral arrow), the lavender OH⁻ pan is forced down (fewer hydroxides). The pans can never both rise or both fall, because their product is pinned to $K_w$. Every worked example on this page is just one tilt of this beam:

• Cell A tilts coral up (acidic) → lavender drops.
• Cell B tilts lavender up (basic) → coral drops.
• Cell C/D is the perfectly level beam — both pans at the same height (equal ions), which is what "neutral" means.

Keep this beam in your head; the algebra below is just numbers for the tilts.

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Cell A — acidic solution

Forecast: $[{\text{H}}^{+}]$ is bigger than $10^{-7}$, so the room is crowded with protons — expect $[{\text{OH}}^{-}]$ to come out smaller than $10^{-7}$, and the label to be acidic. On the seesaw: coral pan pushed up, lavender pan drops.

1. Write the rule. $K_w=[{\text{H}}^{+}][{\text{OH}}^{-}]=1.0\times 10^{-14}$. Why this step? Every Kw problem starts by anchoring to the one constant relationship.
2. Rearrange for the unknown. $[{\text{OH}}^{-}]=\frac{K_w}{[{\text{H}}^{+}]}$. Why this step? We know $[{\text{H}}^{+}]$ and want $[{\text{OH}}^{-}]$; divide the constant by the known one.
3. Substitute. $[{\text{OH}}^{-}]=\frac{1.0\times 10^{-14}}{4.0\times 10^{-4}}=2.5\times 10^{-11}$ M. Why this step? $\frac{1.0}{4.0}=0.25$, and $10^{-14}/10^{-4}=10^{-10}$, so $0.25\times 10^{-10}=2.5\times 10^{-11}$.
4. Label it. $[{\text{H}}^{+}]=4.0\times 10^{-4}>[{\text{OH}}^{-}]=2.5\times 10^{-11}$, so acidic. Why this step? Acidic means more protons than hydroxides.

Verify: multiply back — $(4.0\times 10^{-4})(2.5\times 10^{-11})=10.0\times 10^{-15}=1.0\times 10^{-14}$ ✓. Powers-of-ten check: $-4+(-11)=-15$, and $4.0\times 2.5=10.0$ carries one power up to $-14$, matching $K_w$ (which we treat as the unitless number above). Forecast confirmed.

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Cell B — basic solution

Forecast: hydroxide is large, so protons must be tiny — far below $10^{-7}$. Expect basic. On the seesaw: lavender pan pushed up, coral pan drops.

1. Anchor: $K_w=[{\text{H}}^{+}][{\text{OH}}^{-}]=1.0\times 10^{-14}$. Why this step? Same locked product.
2. Solve for the missing proton concentration: $[{\text{H}}^{+}]=\frac{K_w}{[{\text{OH}}^{-}]}$. Why this step? This time $[{\text{OH}}^{-}]$ is known, so it goes in the denominator.
3. Substitute: $[{\text{H}}^{+}]=\frac{1.0\times 10^{-14}}{4.0\times 10^{-3}}=2.5\times 10^{-12}$ M. Why this step? $\frac{1.0}{4.0}=0.25$, and $10^{-14}/10^{-3}=10^{-11}$, giving $0.25\times 10^{-11}=2.5\times 10^{-12}$.
4. Label: $[{\text{H}}^{+}]=2.5\times 10^{-12}<[{\text{OH}}^{-}]$, so basic.

Verify: $(2.5\times 10^{-12})(4.0\times 10^{-3})=10.0\times 10^{-15}=1.0\times 10^{-14}$ ✓.

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Cell C — exactly neutral at 25 °C

Forecast: with nothing added, the two ions are born in pairs (one H⁺ and one OH⁻ per broken water), so they must be equal — and their product is $10^{-14}$, so each should be $10^{-7}$. On the seesaw: the beam sits perfectly level.

1. State the equality. In pure water $[{\text{H}}^{+}]=[{\text{OH}}^{-}]$; call this common value $x$. Why this step? The autoionization ${\text{H}}_2\text{O}\rightleftharpoons {\text{H}}^{+}+{\text{OH}}^{-}$ makes one of each, so they rise together.
2. Substitute into the rule: $K_w=x\cdot x=x^2$. Why this step? Replacing both factors by the same $x$ turns the product into a square — that is why a square root is the right tool here (it undoes the squaring).
3. Solve: $x=\sqrt{1.0\times 10^{-14}}=1.0\times 10^{-7}$ M. Why this step? $\sqrt{10^{-14}}=10^{-7}$ because half of the exponent $-14$ is $-7$.

Verify: $(1.0\times 10^{-7})(1.0\times 10^{-7})=1.0\times 10^{-14}$ ✓; and pH $=-\log (10^{-7})=7$, the familiar neutral value. Forecast confirmed. See pH and pOH scale.

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Cell D — neutral, but NOT at 25 °C

Forecast: hotter water ionizes more (autoionization is endothermic — see Le Chatelier's principle), so more of both ions. $[{\text{H}}^{+}]$ should be bigger than $10^{-7}$, so pH should drop below 7 — yet the water is still perfectly neutral. On the seesaw: still level, but both pans loaded heavier.

1. Neutral still means equal: $[{\text{H}}^{+}]=[{\text{OH}}^{-}]=x$, so $K_w=x^2$. Why this step? "Neutral" is defined by equality of the two ions, not by pH = 7. Nothing added, so pairs are still equal.
2. Solve: $x=\sqrt{9.6\times 10^{-14}}=3.10\times 10^{-7}$ M. Why this step? $\sqrt{9.6}=3.10$ and $\sqrt{10^{-14}}=10^{-7}$.
3. pH: $\text{pH}=-{\log }_{10}(3.10\times 10^{-7})=6.51$. Why this step? $-\log (3.10\times 10^{-7})=7-\log 3.10=7-0.49=6.51$.
4. Interpret. pH $=6.51<7$, but $[{\text{H}}^{+}]=[{\text{OH}}^{-}]$, so it is neutral. Why this step? This is the whole trap of Cell D: pH 7 is only the neutral point at 25 °C.

Verify: $(3.10\times 10^{-7}{)}^2=9.61\times 10^{-14}\approx 9.6\times 10^{-14}$ ✓.

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Cell E — the degenerate limit: super-dilute acid

Here is the case beginners get wrong. What is the pH of $1.0\times 10^{-8}$ M HCl? The naive answer "pH = 8" is impossible — adding acid to water cannot make it basic! The trap: when the added acid is as tiny as water's own ionization ($\sim 10^{-7}$), you cannot ignore the H⁺ that water itself produces.

Read the figure before the algebra. The dashed coral curve is the naive answer pH $=-\log (\text{added acid})$; follow it left and it crosses pH 7 and climbs into basic territory — which is nonsense for an acid. The solid lavender curve is the true answer from the quadratic below; notice how it bends and flattens toward pH 7 as the acid gets more dilute, never crossing above it. The mint dotted line marks pure-water pH 7. The slate dot at $\log =-8$ is exactly the problem we solve here: the true curve sits at pH 6.98, just below 7, while the naive dashed curve wrongly reads pH 8. The gap between the two curves is the H⁺ that water itself contributes.

Forecast: the acid is so dilute it barely beats pure water. Expect pH just slightly below 7 (around 6.98), NOT 8.

1. Name every source of H⁺. From the strong acid (Strong acids and bases): a fixed $10^{-8}$ M. From water: an extra amount $y$ of H⁺, and because water makes ions in pairs, that same water contributes $y$ of OH⁻ too. So $[{\text{OH}}^{-}]=y$. Why this step? At normal concentrations we drop water's contribution; at $10^{-8}$ M it is comparable, so it must stay.
2. Define the total. Let $h=[{\text{H}}^{+}]=10^{-8}+y$ — the acid's H⁺ plus water's H⁺. Since $[{\text{OH}}^{-}]=y=h-10^{-8}$, we can write everything in one symbol $h$. Why this step? Collapsing $y$ into $h$ removes the auxiliary variable so we have one unknown, not two.
3. Charge balance (the room must be electrically neutral): total positive = total negative. $[{\text{H}}^{+}]=[{\text{Cl}}^{-}]+[{\text{OH}}^{-}]=10^{-8}+[{\text{OH}}^{-}].$ Why this step? Cl⁻ comes only from the acid ($10^{-8}$ M); OH⁻ comes only from water. Positives must match negatives — and this is exactly $h=10^{-8}+y$ from step 2.
4. Use $[{\text{OH}}^{-}]=K_w/[{\text{H}}^{+}]$ to eliminate OH⁻: $h=10^{-8}+\frac{10^{-14}}{h}.$ Why this step? This links the two ions through the locked product, giving one equation in one unknown $h$.
5. Multiply by $h$ and rearrange into a quadratic: $h^2-10^{-8}h-10^{-14}=0.$ Why this step? A quadratic is the honest tool once we refuse to throw away water's ions.
6. Solve with the quadratic formula (keep the positive root): $h=\frac{10^{-8}+\sqrt{(10^{-8}{)}^2+4\times 10^{-14}}}{2}=1.05\times 10^{-7}\text{M}.$ Why this step? Concentrations can't be negative, so we take the $+$ root.
7. pH $=-\log (1.05\times 10^{-7})=6.98$. Why this step? pH is defined as $-{\log }_{10}[{\text{H}}^{+}]$, and $h$ is that total H⁺; taking the log converts the concentration into the familiar 0–14 scale.

Verify: plug back — $[{\text{OH}}^{-}]=10^{-14}/(1.05\times 10^{-7})=9.5\times 10^{-8}$; and $[{\text{H}}^{+}]-[{\text{OH}}^{-}]=1.05\times 10^{-7}-9.5\times 10^{-8}=1.0\times 10^{-8}=[{\text{Cl}}^{-}]$ ✓. pH 6.98 < 7 (acidic), never 8. Trap avoided.

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Cell E' — the mirror: super-dilute base

Everything in Cell E has a perfect reflection: dump a tiny amount of strong base into water and the same trap appears, flipped. The naive "pH = 6" for $10^{-8}$ M NaOH is impossible — adding base cannot make water acidic!

Forecast: by symmetry with Cell E, expect pH just slightly above 7 (around 7.02), NOT 6.

1. Name every source of OH⁻. From the strong base (Strong acids and bases): a fixed $10^{-8}$ M. From water: an extra $y$ of OH⁻ and the same $y$ of H⁺ (paired). So $[{\text{H}}^{+}]=y$. Why this step? Same reasoning as Cell E, mirrored: near $10^{-7}$ water's own ions matter.
2. Define the total. Let $b=[{\text{OH}}^{-}]=10^{-8}+y$, so $[{\text{H}}^{+}]=y=b-10^{-8}$. One symbol $b$ now carries everything. Why this step? Removing the auxiliary $y$ leaves one unknown.
3. Charge balance: total positive = total negative. $[{\text{Na}}^{+}]+[{\text{H}}^{+}]=[{\text{OH}}^{-}],\text{i.e.}10^{-8}+[{\text{H}}^{+}]=b.$ Why this step? Na⁺ ($10^{-8}$ M) and H⁺ are the positives; OH⁻ is the negative. This is $b=10^{-8}+y$ from step 2.
4. Use $[{\text{H}}^{+}]=K_w/[{\text{OH}}^{-}]$ to eliminate H⁺: $b=10^{-8}+\frac{10^{-14}}{b}.$ Why this step? The locked product links the ions, giving one equation in $b$ — identical shape to Cell E.
5. Rearrange into a quadratic and take the positive root: $b^2-10^{-8}b-10^{-14}=0\Rightarrow b=1.05\times 10^{-7}\text{M}.$ Why this step? Same maths as Cell E, so the same number for the abundant ion — here it is OH⁻.
6. Convert to pH via $[{\text{H}}^{+}]=K_w/b$, then pH: $[{\text{H}}^{+}]=\frac{10^{-14}}{1.05\times 10^{-7}}=9.5\times 10^{-8}\text{M},\text{pH}=-\log (9.5\times 10^{-8})=\mathrm{7.02.}$ Why this step? pH is defined on H⁺; the base gave us OH⁻, so we cross back through $K_w$ first.

Verify: $[{\text{OH}}^{-}]-[{\text{H}}^{+}]=1.05\times 10^{-7}-9.5\times 10^{-8}=1.0\times 10^{-8}=[{\text{Na}}^{+}]$ ✓, and pH 7.02 > 7 (basic), never 6. The mirror of Cell E holds exactly.

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Cell F — pH to pOH conversion

Forecast: low pOH means lots of OH⁻ → basic, so pH should be high (near 11–12).

1. Use $\text{pH}+\text{pOH}=14$ (valid at 25 °C). Why this step? Taking $-\log$ of $K_w=[{\text{H}}^{+}][{\text{OH}}^{-}]$ turns the product into a sum: $\text{p}{K}_w=\text{pH}+\text{pOH}$, and $\text{p}{K}_w=-\log (10^{-14})=14$.
2. Solve: $\text{pH}=14-2.4=11.6$. Why this step? Simple subtraction — that is the whole payoff of the log trick.
3. Convert to $[{\text{H}}^{+}]$: $[{\text{H}}^{+}]=10^{-\text{pH}}=10^{-11.6}=2.5\times 10^{-12}$ M. Why this step? $10^{-\text{pH}}$ undoes $-\log$; it's the inverse operation.

Verify: $\text{pOH}=2.4\Rightarrow [{\text{OH}}^{-}]=10^{-2.4}=4.0\times 10^{-3}$; product $(2.5\times 10^{-12})(4.0\times 10^{-3})=1.0\times 10^{-14}$ ✓. See pH and pOH scale.

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Cell G — real-world word problem

Forecast: pH 2.3 is strongly acidic, so hydroxide will be brutally scarce — expect around $10^{-12}$ M.

1. Translate pH to $[{\text{H}}^{+}]$: $[{\text{H}}^{+}]=10^{-2.30}=5.0\times 10^{-3}$ M. Why this step? The recipe speaks in pH; the Kw tool speaks in concentration, so convert first.
2. Apply the rule: $[{\text{OH}}^{-}]=\frac{K_w}{[{\text{H}}^{+}]}=\frac{1.0\times 10^{-14}}{5.0\times 10^{-3}}$. Why this step? The one relationship that connects the two ions.
3. Compute: $[{\text{OH}}^{-}]=2.0\times 10^{-12}$ M. Why this step? $\frac{1.0}{5.0}=0.20$, $10^{-14}/10^{-3}=10^{-11}$, product $2.0\times 10^{-12}$.

Verify: $(5.0\times 10^{-3})(2.0\times 10^{-12})=1.0\times 10^{-14}$ ✓. Per litre that's $2.0\times 10^{-12}\times 6.02\times 10^{23}\approx 1.2\times 10^{12}$ hydroxide ions — a trillion, yet chemically negligible. Forecast confirmed.

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Cell H — exam twist: pOH given at a non-25 °C temperature

Forecast: the trap is to blurt "pH = 9." But $\text{p}{K}_w\ne 14$ here, so recompute the sum first.

1. Find $\text{p}{K}_w$: $\text{p}{K}_w=-{\log }_{10}(2.9\times 10^{-14})=13.54$. Why this step? The pH–pOH sum equals $\text{p}{K}_w$, and $\text{p}{K}_w$ shifts with temperature because $K_w$ does.
2. Use the corrected identity: $\text{pH}+\text{pOH}=\text{p}{K}_w=13.54$. Why this step? This is the general form; "14" was only the 25 °C special case.
3. Solve: $\text{pH}=13.54-5.00=8.54$. Why this step? Subtraction, exactly as in Cell F but with the right total.

Verify: $[{\text{OH}}^{-}]=10^{-5.00}=1.0\times 10^{-5}$; $[{\text{H}}^{+}]=10^{-8.54}=2.9\times 10^{-9}$; product $(1.0\times 10^{-5})(2.9\times 10^{-9})=2.9\times 10^{-14}=K_w$ ✓. Using 14 would have given 9.00 — wrong by 0.46 pH units.

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The whole map at a glance

The flowchart below uses the same names as the worked steps: the node "given H+" means "you were handed $[{\text{H}}^{+}]$" (as in Cell A), "given OH-" means "you were handed $[{\text{OH}}^{-}]$" (Cell B), and "Kw / h" / "Kw / OH" are the exact divisions from step 2 of those examples. Follow the arrows the way you would triage a real problem.

Recall Self-test: name the cell

Given only $[{\text{OH}}^{-}]$ at 25 °C, which tool? ::: Divide: $[{\text{H}}^{+}]=K_w/[{\text{OH}}^{-}]$ (Cell B). Pure water at 90 °C — is pH 7? ::: No. Neutral means $[{\text{H}}^{+}]=[{\text{OH}}^{-}]$; pH is below 7 because $K_w>10^{-14}$ (Cell D). Why does $10^{-8}$ M HCl need a quadratic? ::: Because the added H⁺ is comparable to water's own $10^{-7}$ M, so water's ions cannot be dropped (Cell E). What is the pH of $10^{-8}$ M NaOH — 6 or just above 7? ::: Just above 7 (pH 7.02); base cannot make water acidic (Cell E'). At 40 °C with $K_w=2.9\times 10^{-14}$, what is pH + pOH? ::: $\text{p}{K}_w=13.54$, not 14 (Cell H). What does the "p" in pOH mean? ::: Take $-{\log }_{10}$; so $\text{pOH}=-{\log }_{10}[{\text{OH}}^{-}]$.

Connections

• Parent: Ionic product of water — the rule these examples exercise
• pH and pOH scale — the log conversions in Cells C, F, H
• Strong acids and bases — full ionization used in Cells E, E', G
• Le Chatelier's principle — why $K_w$ rises with temperature (Cells D, H)
• Buffer solutions — next level: resisting these shifts
• Common ion effect — suppressing water's ionization
• Weak acids Ka — pairs with $K_w=K_a\times K_b$
• Solubility product Ksp — same "locked product" logic
• Arrhenius theory — defines the acidic/basic labels used throughout