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

2.6.9 · Chemistry › Equilibrium

The autoionization equilibrium

Yeh hota kyun hai? Water polar hoti hai. Ek molecule ka δ⁺ hydrogen doosri molecule ke δ⁻ oxygen ki taraf attract hota hai. Kabhi-kabhi, thermal energy ek complete proton transfer karati hai. Yeh rare hai — har 10⁹ water molecules mein se sirf lagbhag 2 hi kisi bhi pal ionized hoti hain — lekin yeh hamesha hota rehta hai.

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

Deriving Kw from first principles

Step 1: Equilibrium expression likho

Kisi bhi equilibrium ke liye, hum likhte hain: $K_{c} = \frac{[ products ]}{[ reactants ]}$

Water autoionization ke liye: $K_{c} = \frac{[ H ^{+} ] [ OH ^{-} ]}{[ H _{2} O ] ^{2}}$

Yeh form kyun? Har concentration term ko uske stoichiometric coefficient ke barabar exponent milta hai. Water left side par do baar aati hai (as reactant), isliye denominator mein squared hai.

Step 2: Samjho ki [H₂O] essentially constant hai

Pure water ki molarity hoti hai: $[H_{2} O] = \frac{1000 g/L}{18 g/mol} = 55.5 M$

Hum ise constant kyun treat kar sakte hain? Jab sirf 10⁻⁷ M water ionize hoti hai, toh hum 55.5 M se 55.4999 M par aa jaate hain — ek negligible change. Liquid water ki concentration ionization ke dauran meaningfully nahi badlati.

Step 3: Constant ko ek naye equilibrium constant mein absorb karo

Kyunki [H₂O] constant hai: $K_{c} \times [H_{2} O]^{2} = [H^{+}] [OH^{-}]$

Kw (ionic product of water) define karo: $K_{w} = [H^{+}] [OH^{-}]$

jahan $K_{w} = K_{c} \times (55.5)^{2}$

Step 4: 25 °C par Kw experimentally determine karo

Conductivity measurements se pata chalta hai ki pure water at 25 °C mein:

$[H^{+}] = 1.0 \times 1 0^{- 7}$ M
$[OH^{-}] = 1.0 \times 1 0^{- 7}$ M

Isliye: $K_{w} = (1.0 \times 1 0^{- 7}) (1.0 \times 1 0^{- 7}) = 1.0 \times 1 0^{- 14} at 25° C$

Critical insight: Yeh relationship KISI BHI aqueous solution ke liye valid hai, sirf pure water ke liye nahin. Agar aap acid add karke [H⁺] badhao, toh [OH⁻] product ko 10⁻¹⁴ par rakhne ke liye zaroor ghategga. Agar aap base add karke [OH⁻] badhao, toh [H⁺] proportionally ghategga.

Kw temperature par kyun depend karta hai

Le Chatelier's principle ke anusaar: Temperature badhane se equilibrium products ki taraf shift hota hai (zyada ions bante hain). Isliye, Kw temperature ke saath badhta hai.

Temperature	Kw value
0 °C	1.14 × 10⁻¹⁵
25 °C	1.00 × 10⁻¹⁴
50 °C	5.47 × 10⁻¹⁴
100 °C	5.13 × 10⁻¹³

Key point: 50 °C par, neutral water mein [H⁺] = 2.34 × 10⁻⁷ M (pH = 6.63) hota hai, lekin yeh phir bhi neutral hai kyunki [H⁺] = [OH⁻]. Neutral ka matlab hamesha pH = 7 nahin hota!

Worked examples

Solution: $K_{w} = [H^{+}] [OH^{-}] = 1.0 \times 1 0^{- 14}$ use karo

$[OH^{-}] = \frac{K _{w}}{[ H ^{+} ]} = \frac{1.0 \times 1 0 ^{- 14}}{1.0 \times 1 0 ^{- 3}} = 1.0 \times 1 0^{- 11} M$

Yeh step kyun? Hum Kw expression ko algebraically rearrange karte hain. Product constant rehna chahiye, isliye agar [H⁺] bada hai (acidic), toh [OH⁻] proportionally chota hona chahiye.

Check: $(1 0^{- 3}) (1 0^{- 11}) = 1 0^{- 14}$ ✓

Solution: $[H^{+}] = \frac{K _{w}}{[ OH ^{-} ]} = \frac{1.0 \times 1 0 ^{- 14}}{1.0 \times 1 0 ^{- 3}} = 1.0 \times 1 0^{- 11} M$

Yeh step kyun? Basic solutions mein [OH⁻] bada hota hai, isliye 10⁻¹⁴ ka product maintain karne ke liye [H⁺] chota hona chahiye. Solution basic hai (pH = 11), lekin H⁺ ions phir bhi exist karte hain — bas bahut dilute hain.

Solution: Pure water mein, $[H^{+}] = [OH^{-}]$ . Is value ko $x$ maano.

$K_{w} = x \cdot x = x^{2}$ $5.47 \times 1 0^{- 14} = x^{2}$ $x = 5.47 \times 1 0^{- 14} = 2.34 \times 1 0^{- 7} M$

Yeh step kyun? Autoionization equation dikhata hai ki ek H⁺ aur ek OH⁻ saath bante hain, isliye pure water mein unki concentrations equal honi chahiye.

pH calculation: $pH = - lo g (2.34 \times 1 0^{- 7}) = 6.63$

Kya yeh acidic hai? Nahin! Yeh 50 °C par neutral hai kyunki [H⁺] = [OH⁻]. Neutrality ka pH temperature ke saath shift karta hai.

Solution: HCl ek strong acid hai: $HCl \to H^{+} + Cl^{-}$

Added H⁺ water equilibrium ko shift karta hai: $H^{+} + OH^{-} \to H_{2} O$

Yeh OH⁻ ions consume karta hai. Naya [OH⁻] hai: $[OH^{-}] = \frac{1 0 ^{- 14}}{0.01} = 1 0^{- 12} M$

Verification:

> K_w = (0.01)(10^{-12}) = 10^{-14} > $$-\log K_w = -\log[\ce{H+}] - \log[\ce{OH-}]$$ $$\text{p}K_w = \text{pH} + \text{pOH}$$At 25 °C: \text{p}K_w = 14, so:$$\boxed{\text{pH} + \text{pOH} = 14}$$ **Yeh useful kyun hai:** Tum pH aur pOH ke beech instantly convert kar sakte ho. Agar pH = 3 hai, toh pOH = 11 hai. Agar [OH⁻] pata hai, toh pOH calculate karo, phir pH paane ke liye 14 mein se ghata do. > [!recall]- Ek 12-saal ke bachche ko samjhao > Ek bheed bhari room imagine karo jahan log constantly haath milaate aur chhodte rehte hain. Water molecules bhi aisi hi hain — yeh hamesha ek doosre se takraati rehti hain, aur kabhi-kabhi ek doosre se ek tiny positive tukda (ek proton) cheen leti hai. Jab aisa hota hai, toh ek molecule ko extra positive charge milta hai (H⁺) aur doosre ko extra negative charge (OH⁻). Yeh zyada der alag nahin rehte — yeh usually ek doosre ko dhundh lete hain aur normal water mein wapas mil jaate hain. Lekin kisi bhi pal, kuch pairs hamesha alag hote hain. 10⁻¹⁴ number humein batata hai ki kitne alag pairs exist karte hain. Yeh kuch aisa kehne jaisa hai ki "1 crore haath milaane wali bheed mein, sirf 1 pair kisi bhi pal alag hai." Itna rare hai! Yahan magic hai: agar tum nimbu ka ras (acid) milao, toh tum room mein tons of H⁺ dump kar rahe ho. Ab itne zyada H⁺ ions hain ki yeh jaldi se saare OH⁻ ions ko pakad lete hain aur combine ho jaate hain. Free OH⁻ ions ki number gir jaati hai. Lekin rule wahi rehta hai — H⁺ ki number times OH⁻ ki number hamesha 10⁻¹⁴ ke barabar hoti hai. Yeh ek seesaw jaisa hai: jab ek side upar jaaye, doosri side *zaroor* niche jaaye taaki balance bana rahe. > [!mnemonic] Yaad rakho Kw = 10⁻¹⁴ > **"K-water is won (1) over ten thousand trillion (10¹⁴)"** Ya socho: **K**w at **25**°C = **1**0⁻**14** → "Quarter-century (25) tumhe 14 zeros wala 1 deta hai" pH + pOH = 14 ke liye: "**P**ositive **H** plus **P**ositive **OH** equals **P**erfect **14**" ## Connections - [[Le Chatelier's principle]] — explain karta hai ki Kw temperature ke saath kyun badhta hai - [[pH and pOH scale]] — dono Kw se derive hote hain - [[Strong acids and bases]] — completely ionize hote hain, water equilibrium shift karte hain - [[Buffer solutions]] — conjugate base concentrations calculate karne ke liye Kw use karte hain - [[Common ion effect]] — ions add karne se water ionization suppress hoti hai - [[Solubility product Ksp]] — ionic solid equilibria ke liye similar logic - [[Arrhenius theory]] — acids/bases ko [H⁺] aur [OH⁻] par effect se define karta hai - [[Weak acids Ka]] — Kw use karo Ka se Kb nikalne ke liye: $K_w = K_a \times K_b$ --- #flashcards/chemistry Water ka ionic product Kw 25 °C par kya hota hai? :: Kw = 1.0 × 10⁻¹⁴ = [H⁺][OH⁻] Water ke autoionization ki equation likho :: H₂O(l) ⇌ H⁺(aq) + OH⁻(aq) YA 2H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq) Agar [H⁺] = 10⁻³ M at 25 °C hai, toh [OH⁻] kya hai? ::: [OH⁻] = Kw/[H⁺] = 10⁻¹⁴/10⁻³ = 10⁻¹¹ M Kw ko constant kyun treat kiya jaata hai jabki ionization ke dauran water concentration ghatti hai? ::: Kyunki [H₂O] ≈ 55.5 M hai aur sirf ~10⁻⁷ M ionize hoti hai, change negligible hai (55.5 vs 55.4999999 M). Essentially constant [H₂O]² term ko Kw mein absorb kar lete hain. Temperature Kw ko kaise affect karta hai aur kyun? ::: Kw temperature ke saath badhta hai kyunki autoionization endothermic hai (ΔH° > 0). Le Chatelier's principle ke anusaar, heat equilibrium ko products ki taraf shift karti hai (zyada H⁺ aur OH⁻). 25 °C par pH aur pOH ke beech kya relationship hai? :: pH + pOH = 14 (kyunki is temperature par pKw = 14 hota hai) Pure water at 50 °C mein Kw = 5.47 × 10⁻¹⁴ hai. [H⁺] nikalo :: [H⁺] = [OH⁻] = √Kw = √(5.47 × 10⁻¹⁴) = 2.34 × 10⁻⁷ M (pH = 6.63, phir bhi neutral!) True ya false: HCl milane se Kw ki value badh jaati hai :: False. Kw sirf temperature par depend karta hai. HCl milane se [H⁺] badhta hai lekin [OH⁻] ghatta hai taaki [H⁺][OH⁻] 10⁻¹⁴ par constant rahe. pH 7 hamesha neutral kyun nahin hota? ::: Neutral ka matlab hai [H⁺] = [OH⁻], jo tab hota hai jab [H⁺] = √Kw. Kyunki Kw temperature ke saath badalta hai, neutral pH bhi badlata hai. 25 °C par neutral pH 7 hai, lekin 50 °C par neutral pH 6.63 hai. Agar [OH⁻] = 1.0 × 10⁻⁴ M hai, toh [H⁺] calculate karo aur batao solution acidic hai ya basic ::: [H⁺] = 10⁻¹⁴/10⁻⁴ = 10⁻¹⁰ M. Kyunki [OH⁻] > [H⁺] hai, solution basic hai (pH = 10, pOH = 4). Pure water ki molarity kya hai aur ise constant kyun treat kar sakte hain? ::: [H₂O] = 1000 g/L ÷ 18 g/mol = 55.5 M. Yeh essentially constant hai kyunki jo thodi si amount ionize hoti hai (~10⁻⁷ M) woh 55.5 M ke comparison mein negligible hai. Conjugate acid-base pair ke liye Kw ko Ka aur Kb mein express karo ::: Kw = Ka × Kb (ek given temperature par) ## 🖼️ Concept Map ```mermaid flowchart TD W[Water is polar] -->|enables proton transfer| A[Autoionization equilibrium] A -->|H2O gives H plus and OH minus| KC[Equilibrium expression Kc] KC -->|H2O conc 55.5 M constant| ABS[Absorb H2O into constant] ABS -->|defines| KW[Kw = H plus times OH minus] KW -->|measured at 25 C| VAL[Kw = 1.0e-14] VAL -->|from| CONC[H plus = OH minus = 1.0e-7 M] KW -->|holds for| AQ[Any aqueous solution] AQ -->|add acid raises H plus| INV[OH minus decreases inversely] KW -->|controls| PH[pH scale 0 to 14] ENDO[Endothermic +56 kJ/mol] -->|Le Chatelier| TEMP[Kw rises with temperature] TEMP -->|changes| VAL ```