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

2.6.16 · D4 · Chemistry › Equilibrium › Salt hydrolysis — pH of salt solutions (4 cases - SA - SB,

Yeh page ek self-test ladder hai. Har problem parent note Salt hydrolysis (4 cases) par build karti hai. Pehle har problem khud try karo, phir collapsible solution kholna.

Shuru karne se pehle, yeh raha toolbox — ek jagah par saare chaar master formulas.

Har letter ka matlab, bilkul scratch se:

$K_{w} = 1 0^{- 14}$ water ka self-splitting constant hai; $p K_{w} = 14$ .
$K_{a}$ measure karta hai ki ek weak acid kitni willingly $H^{+}$ deti hai — bada $K_{a}$ = zyada strong acid.
$K_{b}$ measure karta hai ki ek weak base kitni willingly $H^{+}$ pakadti hai — bada $K_{b}$ = zyada strong base.
$lo g$ yahan base-10 hai. $lo g (1 0^{- n}) = - n$ . Bas yahi ek log fact chahiye.

Level 1 — Recognition

(Kya tum case name kar sakte ho aur pH ki direction predict kar sakte ho?)

Exercise 1.1

Har salt ko SA–SB, SA–WB, WA–SB, ya WA–WB classify karo, aur batao ki iska solution acidic, basic, ya neutral hai: (a) $KNO_{3}$ (b) $NH_{4} Cl$ (c) $CH_{3} COONa$ (d) $NH_{4} CN$ .

Recall Solution 1.1

Har salt ko uske parents mein split karo (acid anion deti hai, base cation deta hai).

(a) $KNO_{3}$ → $K^{+}$ ( $KOH$ se, strong base) + $NO_{3}^{-}$ ( $HNO_{3}$ se, strong acid) = SA–SB → neutral, pH = 7.
(b) $NH_{4} Cl$ → $NH_{4}^{+}$ ( $NH_{3}$ se, weak base) + $Cl^{-}$ ( $HCl$ se, strong acid) = SA–WB → acidic, pH < 7.
(c) $CH_{3} COONa$ → $Na^{+}$ ( $NaOH$ se, strong base) + $CH_{3} COO^{-}$ ( $CH_{3} COOH$ se, weak acid) = WA–SB → basic, pH > 7.
(d) $NH_{4} CN$ → $NH_{4}^{+}$ (weak base parent) + $CN^{-}$ ( $HCN$ se, weak acid) = WA–WB → $K_{a}$ vs $K_{b}$ par depend karta hai.

Exercise 1.2

Bina kisi calculator ke, in solutions ko (sab 0.1 M) increasing pH ke order mein arrange karo: $HCl$ ki salt $NaCl$ , $NH_{4} Cl$ , $CH_{3} COONa$ .

Recall Solution 1.2

Acidic sabse neeche hoga, neutral beech mein, basic sabse upar.

$NH_{4} Cl$ = acidic → sabse kam pH
$NaCl$ = neutral → pH = 7
$CH_{3} COONa$ = basic → sabse zyada pH

Increasing pH: $NH_{4} Cl < NaCl < CH_{3} COONa$ .

Level 2 — Application

(Sahi formula mein plug karo aur number nikalo.)

Exercise 2.1

0.1 M $NH_{4} Cl$ solution ka pH nikalo. Diya gaya: $K_{b} (NH_{3}) = 1.8 \times 1 0^{- 5}$ , $K_{w} = 1 0^{- 14}$ .

Recall Solution 2.1

Case: SA–WB (acidic). $pH = \frac{1}{2} (p K_{w} - p K_{b} - lo g C)$ use karo.

$p K_{w} = 14$ .
$p K_{b} = - lo g (1.8 \times 1 0^{- 5}) = 4.74$ .
$lo g C = lo g (0.1) = - 1$ .

$pH = \frac{1}{2} (14 - 4.74 - (- 1)) = \frac{1}{2} (14 - 4.74 + 1) = \frac{10.26}{2} = 5.13$ pH ≈ 5.13 (acidic ✓ — jaisa ek SA–WB salt hona chahiye).

Exercise 2.2

0.01 M sodium acetate ( $CH_{3} COONa$ ) ka pH nikalo. Diya gaya: $K_{a} (CH_{3} COOH) = 1.8 \times 1 0^{- 5}$ , $K_{w} = 1 0^{- 14}$ .

Recall Solution 2.2

Case: WA–SB (basic). $pH = \frac{1}{2} (p K_{w} + p K_{a} + lo g C)$ use karo.

$p K_{a} = - lo g (1.8 \times 1 0^{- 5}) = 4.74$ .
$lo g C = lo g (0.01) = - 2$ .

$pH = \frac{1}{2} (14 + 4.74 + (- 2)) = \frac{16.74}{2} = 8.37$ pH ≈ 8.37 (basic ✓).

Exercise 2.3

Ammonium acetate ( $CH_{3} COONH_{4}$ ) ka pH kisi bhi concentration par nikalo. Diya gaya: $K_{a} = 1.8 \times 1 0^{- 5}$ , $K_{b} = 1.8 \times 1 0^{- 5}$ .

Recall Solution 2.3

Case: WA–WB. $pH = \frac{1}{2} (p K_{w} + p K_{a} - p K_{b})$ use karo. Kyunki $K_{a} = K_{b}$ hai, $p K_{a} = p K_{b} = 4.74$ : $pH = \frac{1}{2} (14 + 4.74 - 4.74) = \frac{14}{2} = 7.0$ pH = 7.0. Note karo ki $C$ ki zaroorat kabhi nahi padi — WA–WB ke liye concentration poori tarah cancel ho jaati hai.

Level 3 — Analysis

(Reason karo ki inputs change hone par pH kaise badhega ya ghategaa.)

Exercise 3.1

0.1 M $NH_{4} Cl$ solution ko 100-fold dilute karke 0.001 M kar diya jaata hai. Kya pH badhega ya ghategaa, aur kitna? ( $K_{b} = 1.8 \times 1 0^{- 5}$ use karo.)

Recall Solution 3.1

SA–WB formula: $pH = \frac{1}{2} (p K_{w} - p K_{b} - lo g C)$ . Sirf $- lo g C$ term badlega. $C = 0.1$ se $C = 0.001$ par jaane par:

$lo g (0.1) = - 1$ , toh term $= + 1$ ; pH mein contribution $= \frac{1}{2} (+ 1) = + 0.5$ .
$lo g (0.001) = - 3$ , toh term $= + 3$ ; contribution $= \frac{1}{2} (+ 3) = + 1.5$ .

pH mein change $= 1.5 - 0.5 = + 1.0$ . pH 1.0 unit badhta hai (5.13 se 6.13). Physical reading: ek acidic salt ko dilute karna usse neutral ki taraf push karta hai (pH 7), kyunki simply hydrolyse karne wala ion kam ho jaata hai. Yeh 7 ki taraf approach karta hai lekin (is simple model mein) kabhi usse cross nahi karta.

Exercise 3.2

Do 0.1 M basic salts: $CH_{3} COONa$ ( $K_{a} = 1.8 \times 1 0^{- 5}$ ) aur $NaCN$ ( $K_{a} (HCN) = 4.9 \times 1 0^{- 10}$ ). Kis ka zyada pH hai, aur kyun?

Recall Solution 3.2

WA–SB formula: $pH = \frac{1}{2} (p K_{w} + p K_{a} + lo g C)$ . Same $K_{w}$ , same $C$ — sirf $p K_{a}$ alag hai.

Acetate: $p K_{a} = 4.74$ → $pH = \frac{1}{2} (14 + 4.74 - 1) = 8.87$ .
Cyanide: $p K_{a} = - lo g (4.9 \times 1 0^{- 10}) = 9.31$ → $pH = \frac{1}{2} (14 + 9.31 - 1) = 11.16$ .

$NaCN$ ka pH zyada hai (≈ 11.16). Kyun: ek weaker parent acid (chhota $K_{a}$ , bada $p K_{a}$ ) ek stronger conjugate base peeche chhod jaati hai. $HCN$ acetic acid se kaafi zyada weak hai, isliye $CN^{-}$ water se $H^{+}$ zyada aggressively pakadta hai → zyada $OH^{-}$ → zyada pH.

Exercise 3.3

Neeche diye map figure ko dekho. Ek sentence mein explain karo ki WA–WB line flat kyun hai jabki baaki do concentration ke saath slope karti hain.

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

Recall Solution 3.3

SA–WB aur WA–SB mein sirf ek ion hydrolyse karta hai, isliye zyada salt = zyada woh ion $H^{+}$ ya $OH^{-}$ produce karta hai → pH $C$ par depend karta hai (sloped lines). WA–WB mein, dono ions hydrolyse karte hain aur har ek doosre ke effect ko proportionally dilute karta hai; cation-hydrolysis ka $C$ aur anion-hydrolysis ka $C$ algebraically cancel ho jaate hain, chhod ke $pH = \frac{1}{2} (p K_{w} + p K_{a} - p K_{b})$ mein koi $C$ nahi → flat line.

Level 4 — Synthesis

(Ideas combine karo; backward kaam karo; tricky WA–WB handle karo.)

Exercise 4.1

Ek WA–SB salt ke 0.1 M solution ka measured pH = 9.0 hai. Parent acid ka $p K_{a}$ nikalo, aur isse $K_{a}$ nikalo. ( $K_{w} = 1 0^{- 14}$ lo.)

Recall Solution 4.1

WA–SB formula ko backward karo: $pH = \frac{1}{2} (p K_{w} + p K_{a} + lo g C)$ $9.0 = \frac{1}{2} (14 + p K_{a} + lo g 0.1) = \frac{1}{2} (14 + p K_{a} - 1)$ $18 = 13 + p K_{a} \Rightarrow p K_{a} = 5.0$ Phir $K_{a} = 1 0^{- 5.0} = 1.0 \times 1 0^{- 5}$ . Answer: $p K_{a} = 5.0$ , $K_{a} = 1.0 \times 1 0^{- 5}$ .

Exercise 4.2

Predict karo ki $NH_{4} CN$ acidic hai ya basic, phir uska pH compute karo. Diya gaya: $K_{a} (HCN) = 4.9 \times 1 0^{- 10}$ , $K_{b} (NH_{3}) = 1.8 \times 1 0^{- 5}$ , $K_{w} = 1 0^{- 14}$ .

Recall Solution 4.2

Predict: WA–WB. $K_{a}$ vs $K_{b}$ compare karo. Yahan $K_{a} = 4.9 \times 1 0^{- 10}$ (tiny) $<$ $K_{b} = 1.8 \times 1 0^{- 5}$ . Base side jeetta hai → basic, pH > 7. Compute $pH = \frac{1}{2} (p K_{w} + p K_{a} - p K_{b})$ se:

$p K_{a} = - lo g (4.9 \times 1 0^{- 10}) = 9.31$ .
$p K_{b} = - lo g (1.8 \times 1 0^{- 5}) = 4.74$ . $pH = \frac{1}{2} (14 + 9.31 - 4.74) = \frac{1}{2} (18.57) = 9.28$ pH ≈ 9.28 (basic ✓, prediction se match karta hai ki weaker parent acid stronger conjugate base deti hai).

Exercise 4.3

$NH_{4} F$ ke liye: $K_{a} (HF) = 6.6 \times 1 0^{- 4}$ , $K_{b} (NH_{3}) = 1.8 \times 1 0^{- 5}$ . Acidic hai ya basic? pH nikalo.

Recall Solution 4.3

WA–WB phir se. Ab $K_{a} = 6.6 \times 1 0^{- 4} > K_{b} = 1.8 \times 1 0^{- 5}$ : acid side jeetti hai → acidic, pH < 7.

$p K_{a} = - lo g (6.6 \times 1 0^{- 4}) = 3.18$ .
$p K_{b} = - lo g (1.8 \times 1 0^{- 5}) = 4.74$ . $pH = \frac{1}{2} (14 + 3.18 - 4.74) = \frac{1}{2} (12.44) = 6.22$ pH ≈ 6.22 (acidic ✓). Note karo ki yeh sirf thoda 7 se neeche hai kyunki koi bhi parent dramatically doosre se stronger nahi hai.

Level 5 — Mastery

(Multi-step, degenerate cases, aur ek limit.)

Exercise 5.1

0.1 M $NH_{4} Cl$ ( $K_{b} = 1.8 \times 1 0^{- 5}$ ) ke liye degree of hydrolysis $h$ compute karo, aur confirm karo ki parent derivation mein use ki gayi approximation $h ≪ 1$ justified hai.

Recall Solution 5.1

$h$ = woh fraction of ion jo react kar gaya. Parent derivation se, $h = \frac{K _{w}}{K _{b} C}$ . $h = \frac{1 0 ^{- 14}}{( 1.8 \times 1 0 ^{- 5} ) ( 0.1 )} = \frac{1 0 ^{- 14}}{1.8 \times 1 0 ^{- 6}} = 5.56 \times 1 0^{- 9} = 7.45 \times 1 0^{- 5}$ Toh sirf lagbhag $0.0075%$ $NH_{4}^{+}$ hydrolyse karta hai. Kyunki $h = 7.45 \times 1 0^{- 5} ≪ 1$ hai, approximation $C (1 - h) \approx C$ excellent hai. ✓ (Cross-check: $[H^{+}] = C h = 0.1 \times 7.45 \times 1 0^{- 5} = 7.45 \times 1 0^{- 6}$ M → pH $= - lo g (7.45 \times 1 0^{- 6}) = 5.13$ , Ex 2.1 se match karta hai.)

Exercise 5.2 (degenerate input)

0.1 M $NaCl$ solution ka pH kya hai, aur kya hota agar koi naively hamare kisi hydrolysis formula mein $NaCl$ plug kar deta?

Recall Solution 5.2

$NaCl$ SA–SB hai: dono ions spectators hain, koi hydrolysis nahi. pH = 7 exactly (25 °C par), $C$ se independent. WA/SB ya SA/WB formulas simply apply nahi hote — koi weak $K_{a}$ ya $K_{b}$ plug karne ke liye hai hi nahi (ek strong acid ka effectively "infinite" $K_{a}$ hota hai, isliye $p K_{a} \to - \infty$ aur formula toot jaata hai). Correct treatment trivial hai: undisturbed water autoionization deta hai $[H^{+}] = [OH^{-}] = 1 0^{- 7}$ , pH 7. Lesson: pehle case identify karo; SA–SB doosre formulas ka ek limit nahi hai, yeh apna khud ka rule hai.

Exercise 5.3 (limit behaviour)

WA–SB formula use karke, $CH_{3} COONa$ ka $C = 1 0^{- 4}$ M par pH nikalo aur $C \to 0$ ke saath trend par comment karo. ( $K_{a} = 1.8 \times 1 0^{- 5}$ .)

Recall Solution 5.3

$pH = \frac{1}{2} (14 + 4.74 + lo g (1 0^{- 4})) = \frac{1}{2} (14 + 4.74 - 4) = \frac{1}{2} (14.74) = 7.37$ . pH ≈ 7.37. Compare karo: 0.1 M par 8.87 tha, 0.01 M par 8.37 tha, $1 0^{- 4}$ M par sirf 7.37. Trend: jaise jaise $C \to 0$ formula pH ko 7 ki taraf push karta hai — dilution hydrolyse karne wale anion ko khaali kar deta hai isliye solution pure water jaisa ho jaata hai. (Bare formula absurdly chhote $C$ ke liye 7 se neeche slide karta rahega; physically yeh 7 par saturate hota hai kyunki water ka apna $1 0^{- 7}$ dominate karta hai — simple model sirf ~ $1 0^{- 3}$ – $1 0^{- 4}$ M tak hi trustworthy hai.)

Exercise 5.4 (full synthesis)

Ek student $KF$ ( $K_{a} (HF) = 6.6 \times 1 0^{- 4}$ ) ko neutral label kar deta hai. 0.5 M solution ke liye, (a) sahi case batao, (b) true pH compute karo, (c) batao ki woh kitne units galat tha.

Recall Solution 5.4

(a) $KF$ → $K^{+}$ (strong base $KOH$ ) + $F^{-}$ (weak acid $HF$ ) = WA–SB → basic, neutral nahi. (b) $p K_{a} = - lo g (6.6 \times 1 0^{- 4}) = 3.18$ ; $lo g C = lo g (0.5) = - 0.301$ . $pH = \frac{1}{2} (14 + 3.18 + (- 0.301)) = \frac{1}{2} (16.879) = 8.44$ True pH ≈ 8.44. (c) Usne 7.0 assume kiya tha; true value 8.44 hai, isliye woh lagbhag 1.44 pH units galat tha — $[H^{+}]$ mein 27 ka factor se zyada.

Recall Har answer ki one-line recap

1.1 KNO₃ neutral / NH₄Cl acidic / CH₃COONa basic / NH₄CN depends ::: parent-splitting rule se match karta hai 2.1 0.1 M NH₄Cl ka pH ::: 5.13 2.2 0.01 M CH₃COONa ka pH ::: 8.37 2.3 CH₃COONH₄ ka pH ::: 7.0 3.1 NH₄Cl ki 100× dilution ::: pH 1.0 unit badhta hai (5.13 → 6.13) 3.2 zyada pH wali salt ::: NaCN (11.16 vs 8.87) 4.1 pKa back-solve ::: 5.0, Ka = 1.0e-5 4.2 NH₄CN ka pH ::: 9.28 (basic) 4.3 NH₄F ka pH ::: 6.22 (acidic) 5.1 NH₄Cl ke liye h ::: 7.45e-5 5.3 1e-4 M acetate ka pH ::: 7.37 5.4 0.5 M KF ka pH ::: 8.44