Salt hydrolysis — pH of salt solutions (4 cases - SA - SB, SA - WB, WA - SB, WA - WB): Visual walkthrough

2.6.16 · D2 · Chemistry › Equilibrium › Salt hydrolysis — pH of salt solutions (4 cases - SA - SB,

Hum banayenge: $pH = \frac{1}{2} (p K_{w} - p K_{b} - lo g C)$ lekin hum in symbols mein se ek bhi use nahi karenge jab tak hum use draw nahi kar lete.

Step 1 — Beaker mein hai kya cheez?

KYA. Hum NH₄Cl (ammonium chloride) jaisa koi salt pure water mein daalte hain. Salt matlab woh dissolve hote hi completely do alag ions mein toot jaata hai: ek cation NH₄⁺ (positive) aur ek anion Cl⁻ (negative).

KYUN. Kisi bhi chemistry se pehle, hume actors jaanne chahiye. Cation bas ek atom-group hai jisme electrons missing hain isliye usmein + charge hai; anion mein − charge hota hai. Jab tak hum nahi jaante ki kaun se pieces tair rahe hain, kuch aur nahi ho sakta.

PICTURE. Paani mein do ion clouds. Cl⁻ (mint) shant hai — woh ek strong parent acid (HCl) se aaya hai aur react karne ki koi khwaish nahi rakhta. NH₄⁺ (coral) bechain hai — woh ek weak parent base (NH₃) se aaya hai aur apna parent wapas chahta hai.

Step 2 — Woh ek reaction jo matter karti hai

KYA. Sirf bechain ion hi react karta hai. NH₄⁺ ek paani ke molecule ko ek proton (H⁺) deta hai: $bechain cation NH_{4}^{+} + bahut zyada hai H_{2} O ⇌ parent rebuild hua NH_{3} + extra acid bana H_{3} O^{+}$

KYUN. Har symbol ek kaam kar raha hai: NH₄⁺ proton donor hai, H₂O proton receiver hai, NH₃ woh weak parent hai jo rebuild ho gaya (proof ki ion ne use "yaad rakha"), aur H₃O⁺ extra hydronium hai — yahi toh saara point hai, kyunki extra H₃O⁺ matlab zyada acidic. Double arrow $⇌$ matlab yeh poora nahi jaata; yeh ek balance pe settle hota hai.

PICTURE. Ek proton (chhota peela dot) NH₄⁺ se H₂O par jump karta hai, NH₃ peeche chodta hai aur H₃O⁺ banata hai.

Step 3 — "Kitna react karta hai" measure karna: equilibrium constant

KYA. Balance pe, hum ek ratio likhte hain jise hydrolysis constant $K_{h}$ kehte hain: $K_{h} = \frac{[ NH _{3} ] [ H ^{+} ]}{[ NH _{4}^{+} ]}$

KYUN yeh tool. Products over reactants ka ratio kyun? Kyunki equilibrium par forward aur backward jumps equal rate se hote hain, aur yeh nikalta hai ki woh particular fraction hamesha same number par settle hoti hai kisi bhi given ion ke liye — chahe tune kitna bhi salt se shuru kiya ho. Woh fixed number hamara anchor hai. Square brackets $[]$ matlab "concentration of," moles per litre mein measure kiya hua (woh species kitni crowded hai). Paani ko bahar rakha jaata hai kyunki woh woh samundar hai jisme sab kuch tairta hai — practically constant.

PICTURE. Ek tarazu: products (upar ke pans NH₃ aur H⁺) versus reactant (neeche ka pan NH₄⁺). $K_{h}$ tarazu ki reading hai.

Step 4 — Humne $K_{b}$ ka cousin kabhi measure nahi kiya, toh use trade karo

KYA. Hum usually $K_{h}$ directly nahi jaante, lekin hum parent base ki strength $K_{b}$ jaante hain. Woh linked hain: $K_{h} = \frac{K _{w}}{K _{b}}$

KYUN. Har conjugate acid–base pair $K_{a} \cdot K_{b} = K_{w}$ follow karta hai, jahan:

$K_{w} = [H^{+}] [OH^{-}] = 1 0^{- 14}$ paani ka apna chhota split-constant hai (25 °C par ek fixed fact),
$K_{b}$ measure karta hai ki base parent NH₃ kitne strongly protons pakadta hai,
aur $K_{h}$ yahan exactly NH₄⁺ ka acid constant $K_{a}$ hai.

Toh $K_{h} = K_{a} = K_{w} / K_{b}$ . Hum ek unknown ( $K_{h}$ ) ko do known numbers ( $K_{w}$ , $K_{b}$ ) se trade karte hain. Yeh poori derivation mein sabse useful trade hai.

PICTURE. Ek see-saw: $K_{b}$ ko upar push karo (stronger base parent) aur $K_{h}$ neeche jaata hai (kam hydrolysis), kyunki unka product constant $K_{w}$ pe pin hai.

Step 5 — ICE picture se book-keeping

KYA. NH₄⁺ ki concentration $C$ se shuru karo. Maano uska ek fraction $h$ react karta hai ( $h$ = degree of hydrolysis, 0 aur 1 ke beech ka number). Tab:

	NH₄⁺	NH₃	H⁺
start	$C$	$0$	$0$
change	$- C h$	$+ C h$	$+ C h$
end	$C (1 - h)$	$C h$	$C h$

KYUN. Har NH₄⁺ jo react karta hai, hum ek NH₄⁺ khoते hain aur ek NH₃ aur ek H⁺ paate hain — ek andar, ek bahar, isliye teeno "change" rows same $C h$ share karte hain. Kyunki NH₄⁺ weak reactor hai, $h ≪ 1$ , toh $C (1 - h) \approx C$ . Har symbol: $C$ = humne jo crowding daali, $h$ = react karne wala fraction, $C h$ = actually bana H⁺ ka amount.

PICTURE. Height $C$ ka ek bar jisme upar se ek patla sliver $C h$ kaat ke product columns mein shift kar diya.

Step 6 — Boxes ko ratio mein daalo

KYA. "End" row ko $K_{h}$ mein substitute karo: $\frac{K _{w}}{K _{b}} = \frac{( C h ) ( C h )}{C} = C h^{2}$ $h$ ke liye solve karo, phir $[H^{+}] = C h$ ke liye: $[H^{+}] = C h = \frac{K _{w} C}{K _{b}}$

KYUN term-by-term. Numerator $[NH_{3}] [H^{+}] = (C h) (C h) = C^{2} h^{2}$ ; denominator $[NH_{4}^{+}] \approx C$ ; fraction ek $C$ cancel karta hai aur $C h^{2}$ bacha leta hai. Use $K_{w} / K_{b}$ ke equal set karke rearrange karne se woh cheez isolate hoti hai jo hum actually chahte hain — $[H^{+}]$ . Dhyan do zyada salt $C$ se $[H^{+}]$ upar jaata hai (root ke andar), aur stronger base parent $K_{b}$ use neeche push karta hai (denominator mein). Dono physically sense banate hain.

PICTURE. Algebra ek flow ke roop mein: do $C h$ boxes multiply hote hain, ek $C$ divide ho jaata hai, square-root square ko undo karta hai.

Step 7 — pH tak pahunchne ke liye logarithm lo

KYA. pH define hoti hai $pH = - lo g [H^{+}]$ se. Dono sides par $- lo g$ apply karo: $pH = - lo g \frac{K _{w} C}{K _{b}} = - \frac{1}{2} lo g (\frac{K _{w} C}{K _{b}})$ $pH = \frac{1}{2} (p K_{w} - p K_{b} - lo g C)$

KYUN log tool. $1 0^{- 6}$ jaisi concentrations compare karna mushkil hai, isliye pH unhe ek friendly scale par squeeze karta hai jahan 1 ka har step matlab das-guna change hai. $p X = - lo g X$ likhne par: $- lo g K_{w} = p K_{w}$ , $+ lo g K_{b} = - p K_{b}$ (sign flip hoti hai kyunki $K_{b}$ neeche hai), aur $- lo g C$ waise hi rehta hai. $\frac{1}{2}$ seedha Step 6 ke square root se aaya.

PICTURE. Ek number line: raw $[H^{+}]$ powers of ten mein spanning collapse hoke neat 0–14 pH ruler par aa jaata hai; humara answer 7 se neeche land karta hai (acidic), jaisa promise kiya tha.

Step 8 — Edge aur mirror cases (koi gap mat chhodo)

KYA. Bilkul same saat steps baaki teen boxes banate hain — sirf actors change hote hain.

Strong acid + strong base (NaCl): dono ions inert hain, Step 2 reaction bilkul nahi hoti. $K_{h} = 0$ , toh $[H^{+}] = K_{w} = 1 0^{- 7}$ → pH = 7.
Weak acid + strong base (CH₃COONa): ab anion react karta hai, H⁺ ki jagah OH⁻ produce karta hai. Derivation mirror karo → $pH = \frac{1}{2} (p K_{w} + p K_{a} + lo g C)$ , 7 se upar.
Weak acid + weak base (CH₃COONH₄): dono ions react karte hain aur ladte hain. $C$ poora cancel ho jaata hai → $pH = \frac{1}{2} (p K_{w} + p K_{a} - p K_{b})$ , direction set hoti hai is baat se ki $K_{a}$ , $K_{b}$ mein se kaun jeetta hai.

KYUN inhe dikhao. Taaki koi reader koi aisa beaker na dekhe jise humne draw nahi kiya. Degenerate case $h \to 0$ neutral water recover karta hai; "ek side par weak" wale dono cases mirror images hain; "dono weak" wala case tug-of-war hai.

PICTURE. Ek pH ruler par chaar beakers, har ek par mark hai ki kaun sa ion bechain hai.

Ek-picture summary

Ek frame poori chain carry karta hai: bechain ion → proton jump → $K_{h}$ ratio → $K_{w} / K_{b}$ se trade → ICE boxes → $K_{w} C / K_{b}$ → $- lo g$ lo → pH 7 se neeche.

Recall Feynman retelling — kisi dost ko batao

"Tum NH₄Cl paani mein daalte ho. Chloride part ek strong acid se aaya tha, toh woh bas wahan baitha rehta hai, bored. Lekin ammonium part ek weak base se aaya tha, toh woh homesick hai — woh ek paani ka molecule pakad ke apni purani self NH₃ rebuild karne ke liye H⁺ kheench leta hai. Woh free hua H⁺ paani ko thoda acidic bana deta hai. Yeh maapne ke liye ki kitna acidic hai, mein balance ratio $K_{h}$ likhta hoon, lekin mujhe yeh directly pata nahi, toh mein use $K_{w} / K_{b}$ se swap kar leta hoon (ek pair jo mujhe pata hai). Mein simple book-keeping karta hoon: amount $C$ daalte hain, ek chhota fraction $h$ react karta hai, toh $C h$ jaita H⁺ banta hai. Plug in karke aur clean up karne par, H⁺ $K_{w} C / K_{b}$ ke roop mein nikalta hai. $- lo g$ lene se woh ugly chhota number ek tidy pH ban jaata hai, aur woh 7 se neeche land karta hai — acidic. Kaun sa ion homesick hai yeh change karo aur bilkul same story baaki teen cases deti hai."

Reveal-check:

K_{h} = K_{w} / K_{b}

kyun hai aur

K_{w} / K_{a}

kyun nahi?

Kyunki hydrolysing ion NH₄⁺, base NH₃ ka conjugate acid hai, aur conjugate pair ke liye

K_{a} K_{b} = K_{w}

hota hai, toh

K_{h} = K_{a} = K_{w} / K_{b}

pH formula mein

\frac{1}{2}

kahan se aata hai?

[H^{+}] = K_{w} C / K_{b}

mein square root se; square root ka

- lo g

lene par

- \frac{1}{2} lo g

milta hai.

Kaun sa case pH ko concentration

C

se independent rakhta hai?

Weak acid + weak base, kyunki

C

dono competing hydrolyses ke beech cancel ho jaata hai.