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

2.6.16 · D3 · Chemistry › Equilibrium › Salt hydrolysis — pH of salt solutions (4 cases - SA - SB, SA - WB, WA - SB, WA - WB)

Yeh page parent topic ke liye "sab kuch ek saath daalo" wali page hai. Hum koi naya theory nahi seekhenge — hum charon formulas ko stress-test karenge har us tarah ke salt ke against jo exam de sakta hai: neutral wala, acidic wala, basic wala, "depends" wala, aur saath mein sneaky edge cases bhi (bahut dilute solutions, ek strong-ish weak acid ka salt, aur ek word problem).

Kisi bhi formula ko apply karne se pehle, yahan parent page ke charon tools dobara likhe hain taaki yeh page apne aap mein complete rahe. Har symbol clearly explain kiya gaya hai.

Formula Charon pH formulas (aur har letter ka matlab)

$C$ = salt ki concentration moles per litre mein (ions kitni crowded hain).
$K_{w}$ = water-splitting constant $= [H^{+}] [OH^{-}] = 1 0^{- 14}$ at $2 5^{\circ} C$ . To $p K_{w} = - lo g K_{w} = 14$ .
$K_{a}$ = weak acid parent ki strength (bada $K_{a}$ = zyada strong acid). $p K_{a} = - lo g K_{a}$ .
$K_{b}$ = weak base parent ki strength (bada $K_{b}$ = zyada strong base). $p K_{b} = - lo g K_{b}$ .
$pH = - lo g [H^{+}]$ , aur $pOH = - lo g [OH^{-}]$ , jisme $pH + pOH = 14$ .

Salt kis se bana	Naam	Formula
Strong acid + Strong base	neutral	$pH = 7$
Strong acid + Weak base	acidic	$pH = \frac{1}{2} (p K_{w} - p K_{b} - lo g C)$
Weak acid + Strong base	basic	$pH = \frac{1}{2} (p K_{w} + p K_{a} + lo g C)$
Weak acid + Weak base	depends	$pH = \frac{1}{2} (p K_{w} + p K_{a} - p K_{b})$

Scenario matrix

Yeh topic jo bhi problem de sakta hai, use is table ke ek cell ki tarah socho. Hamare kaam mein neeche har cell ko kam se kam ek baar hit karna hai.

#	Cell (scenario)	Expected direction	Example jo isse cover karta hai
A	SA/SB neutral salt	pH = 7 exactly	Ex 1 (KNO₃)
B	SA/WB acidic salt	pH < 7	Ex 2 (NH₄Cl)
C	WA/SB basic salt	pH > 7	Ex 3 (KCN)
D	WA/WB, $K_{a} > K_{b}$	thoda acidic	Ex 4 (NH₄F)
E	WA/WB, $K_{a} < K_{b}$	thoda basic	Ex 5 (NH₄ acetate variant)
F	WA/WB, $K_{a} = K_{b}$	exactly 7	Ex 5b (Ex 5 ke andar check kiya)
G	Degenerate: bahut dilute basic salt	7 ki taraf push	Ex 6 (10⁻⁴ M acetate)
H	Limiting: concentration badh gayi acidic salt ke liye	7 se door push	Ex 7 (1 M NH₄Cl)
I	Word problem / real world	pehle samjho phir compute karo	Ex 8 (soap-water)
J	Exam twist: pH se $K_{a}$ nikalo (reverse)	back-solve	Ex 9

Neeche diya figure har example ke liye haara compass hai: hum pH 7 ke kis taraf land karte hain, aur kyun.

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

Worked example Ex 1 — Cell A:

$0.5 M KNO_{3}$ Forecast: KNO₃ = KOH (strong base) + HNO₃ (strong acid). Dono parents strong hain. Aage padhne se pehle pH guess karo.

Step 1 — Parents identify karo. K⁺ aaya KOH (strong) se, NO₃⁻ aaya HNO₃ (strong) se. Yeh step kyun? Jo formula use karna hai woh sirf parent ki strength se decide hota hai — identification kabhi skip mat karo.

Step 2 — Poocho "kya koi ion paani ke saath react karna chahta hai?" Na K⁺ ka koi weak parent hai jisme wapas jaaye, na NO₃⁻ ka, isliye na koi H⁺ pakadta hai na OH⁻. Yeh step kyun? Hydrolysis tabhi hoti hai jab ion "unsatisfied" ho. Yahan dono satisfied hain.

Step 3 — Neutral rule apply karo. Koi hydrolysis nahi ⇒ paani undisturbed hai ⇒ $[H^{+}] = K_{w} = 1 0^{- 7}$ . $pH = - lo g (1 0^{- 7}) = 7$ Yeh step kyun? Jab koi extra H⁺ ya OH⁻ nahi, toh sirf paani ka apna split pH set karta hai.

Verify: Dekho $C = 0.5$ kabhi calculation mein aaya hi nahi — sahi hai, kyunki neutral salt ka pH concentration-independent hota hai. pH = 7.00. ✓

Worked example Ex 3 — Cell C:

$0.05 M KCN$ Forecast: CN⁻ weak acid HCN se aaya; K⁺ strong KOH se. Anion hydrolyze karega → basic, pH > 7. HCN ek bahut weak acid hai ( $K_{a} = 6.2 \times 1 0^{- 10}$ ), isliye strongly basic solution expect karo.

Given: $C = 0.05$ , $K_{a} (HCN) = 6.2 \times 1 0^{- 10}$ ⇒ $p K_{a} = - lo g (6.2 \times 1 0^{- 10}) = 9.21$ .

Step 1 — WA/SB formula chuno. $pH = \frac{1}{2} (p K_{w} + p K_{a} + lo g C)$ Yeh step kyun? Sirf anion react karta hai; yeh anionic-hydrolysis ka result hai.

Step 2 — Substitute karo. $lo g (0.05) = - 1.30$ . $pH = \frac{1}{2} (14 + 9.21 - 1.30) = \frac{1}{2} (21.91) = 10.95$ Yeh step kyun? Bada $p K_{a}$ (weak acid) hi pH ko itna high push karta hai.

Verify: pH = 10.95, strongly basic — "bahut weak acid parent" forecast se match karta hai. Sanity check: $[OH^{-}] = K_{w} C / K_{a} = 1 0^{- 14} \cdot 0.05/6.2 \times 1 0^{- 10} = 8.98 \times 1 0^{- 4}$ , jisse $pOH = 3.05$ , $pH = 10.95$ . ✓

Worked example Ex 4 — Cell D:

$0.2 M NH_{4} F$ ( $K_{a} > K_{b}$ ) Forecast: Weak acid HF ( $K_{a} = 6.6 \times 1 0^{- 4}$ ) + weak base NH₃ ( $K_{b} = 1.8 \times 1 0^{- 5}$ ). Yahan $K_{a} > K_{b}$ , matlab acid parent stronger hai, isliye uska conjugate anion (F⁻) weaker base hai — cation ka H⁺ production jeet jaata hai. Expect karo slightly acidic, pH < 7.

Given: $p K_{a} = - lo g (6.6 \times 1 0^{- 4}) = 3.18$ , $p K_{b} = - lo g (1.8 \times 1 0^{- 5}) = 4.74$ .

Step 1 — WA/WB formula use karo. $pH = \frac{1}{2} (p K_{w} + p K_{a} - p K_{b})$ Yeh step kyun? Dono ions hydrolyze karte hain; concentration $C$ bilkul cancel ho jaata hai.

Step 2 — Substitute karo. $pH = \frac{1}{2} (14 + 3.18 - 4.74) = \frac{1}{2} (12.44) = 6.22$ Yeh step kyun? Kyunki yahan $p K_{a} < p K_{b}$ hai, middle terms pH ko 7 se neeche le jaate hain.

Verify: pH = 6.22 < 7, " $K_{a} > K_{b}$ ⇒ acidic" se match karta hai. Note karo $C = 0.2$ kabhi matter nahi kiya — WA/WB ki defining feature. ✓

Worked example Ex 6 — Cell G (degenerate: bahut dilute basic salt)

$1 0^{- 4} M$ sodium acetate Forecast: Same acetate jaise parent mein, lekin bahut dilute. Chhote $C$ ke saath $+ lo g C$ term ek bada negative number hai, jo pH ko 7 ki taraf neeche kheenchta hai. Expect karo sirf mildly basic.

Given: $C = 1 0^{- 4}$ , $K_{a} = 1.8 \times 1 0^{- 5}$ ⇒ $p K_{a} = 4.74$ , $lo g C = - 4$ .

Step 1 — WA/SB formula. $pH = \frac{1}{2} (14 + 4.74 - 4) = \frac{1}{2} (14.74) = 7.37$ Yeh step kyun? Dekho $lo g C = - 4$ pH ko neutral ki taraf kheench raha hai jaise dilution badhti hai.

Step 2 — Trend par sanity check. $0.01$ M par parent mein 8.37 aaya tha; $1 0^{- 4}$ M par 7.37 aa raha hai — 100× dilution ne pH ko 1 unit drop kiya, jaisa ki $\frac{1}{2} lo g (100) = 1$ predict karta hai. Yeh step kyun? Confirm karta hai ki formula "compass" figure ki tarah behave kar raha hai: dilution → 7 ki taraf.

Verify: pH = 7.37. (Caveat: $\sim 1 0^{- 5}$ M se neeche, paani ka apna $1 0^{- 7}$ H⁺ matter karne lagta hai aur simple formula fail ho jaata hai — exam mein yeh genuine degenerate limit flag karo.) ✓

Worked example Ex 8 — Cell I (real-world word problem): soapy water

Statement: Ghar ka "soap" mostly sodium stearate hota hai, jo weak acid stearic acid ( $K_{a} = 1.3 \times 1 0^{- 5}$ ) ka sodium salt hai. Tum haath dhone ke liye iska $0.02 M$ solution banaate ho. Kya yeh acidic, neutral, ya basic hai, aur pH kya hai? Forecast: Sodium (strong-base parent) + stearate (weak-acid conjugate) ⇒ WA/SB ⇒ basic (wahi "slippery, slightly alkaline" feel jo soap mein hoti hai). pH ≈ 8 guess karo.

Step 1 — Classify karo. Na⁺ inert hai; stearate⁻ hydrolyze hota hai OH⁻ banane ke liye. Yeh step kyun? Word problems bhi "matrix ke kis cell mein hai?" par reduce ho jaate hain.

Step 2 — WA/SB formula. $p K_{a} = - lo g (1.3 \times 1 0^{- 5}) = 4.89$ , $lo g (0.02) = - 1.70$ . $pH = \frac{1}{2} (14 + 4.89 - 1.70) = \frac{1}{2} (17.19) = 8.60$ Yeh step kyun? "Soap kyun slippery-basic hota hai?" wale sawaal ko ek number mein badal deta hai.

Verify: pH = 8.60, mildly basic — roz ki zindagi ke experience se match karta hai ki soap water gentle-alkaline hota hai, caustic nahi. ✓

Worked example Ex 9 — Cell J (exam twist:

$K_{a}$ reverse-engineer karo) Statement: NaA salt ka $0.1 M$ solution $pH = 8.87$ measure karta hai. Parent acid HA ka $K_{a}$ nikalo. Forecast: pH > 7 ⇒ WA/SB salt (basic), consistent hai. Hum WA/SB formula ko invert karke $p K_{a}$ solve karenge.

Step 1 — WA/SB se start karo aur $p K_{a}$ isolate karo. $pH = \frac{1}{2} (14 + p K_{a} + lo g C) \Rightarrow p K_{a} = 2 pH - 14 - lo g C$ Yeh step kyun? Formula ko algebraically undo karne se measured pH acid constant mein badal jaata hai.

Step 2 — Substitute karo $pH = 8.87$ , $lo g (0.1) = - 1$ . $p K_{a} = 2 (8.87) - 14 - (- 1) = 17.74 - 14 + 1 = 4.74$ $K_{a} = 1 0^{- 4.74} = 1.82 \times 1 0^{- 5}$ Yeh step kyun? Maanga gaya constant deliver karta hai; $\approx 1.8 \times 1 0^{- 5}$ ⇒ HA acetic acid hai.

Verify: Plug back karo: $\frac{1}{2} (14 + 4.74 - 1) = 8.87$ ✓. Toh $K_{a} \approx 1.8 \times 1 0^{- 5}$ . ✓

Recall Quick self-test

$Na_{2} SO_{4}$ kis cell mein aata hai, aur uska pH kya hai? ::: Cell A (SA/SB, H₂SO₄ + NaOH) → pH = 7. WA/WB salt mein concentration kyun nahi aata? ::: Kyunki $C$ do coupled hydrolyses ke beech cancel ho jaata hai; $[H^{+}] = K_{w} K_{a} / K_{b}$ mein koi $C$ nahi hota. Ek basic salt ko 100× dilute karne par pH kitna move hota hai, aur kis taraf? ::: $\frac{1}{2} lo g 100 = 1$ unit neeche, 7 ki taraf. WA/WB salt mein agar $K_{a} > K_{b}$ ho toh acidic hai ya basic? ::: Acidic (H⁺ production jeet jaata hai).