Exercises — Solubility product Ksp — common-ion suppression, selective precipitation

WHAT hum karte hain: har solid ko uske ions mein split karo, phir ion concentrations ko multiply karo, har ek apne stoichiometric number tak raised. WHY: $K_{sp}$ sirf dissolution ka equilibrium constant hai, aur pure solid ki activity $1$ hoti hai to woh kabhi appear nahi hoti.

(a) $\text{AgBr(s)}\rightleftharpoons {\text{Ag}}^{+}+{\text{Br}}^{-}$ $K_{sp}=[{\text{Ag}}^{+}][{\text{Br}}^{-}]$ (b) ${\text{PbCl}}_2\text{(s)}\rightleftharpoons {\text{Pb}}^{2+}+2{\text{Cl}}^{-}$ $K_{sp}=[{\text{Pb}}^{2+}][{\text{Cl}}^{-}{]}^2$ (c) ${\text{Ca}}_3({\text{PO}}_4{)}_2\text{(s)}\rightleftharpoons 3{\text{Ca}}^{2+}+2{\text{PO}}_4^{3-}$ $K_{sp}=[{\text{Ca}}^{2+}{]}^3[{\text{PO}}_4^{3-}{]}^2$

Figure 2 dikhata hai kyun exponents aate hain: har formula unit ek fixed head-count of ions mein toot ta hai (ek, do, ya teen har ek ke), aur woh head-count power ban jaati hai.

"Current reading" $Q$ ko "red line" $K_{sp}$ se compare karo. $Q=5\times 10^{-9}>1.8\times 10^{-10}=K_{sp}$ Kyunki $Q>K_{sp}$, solution supersaturated hai — precipitate forma hai jab tak $Q$ vapas $K_{sp}$ par nahi aa jaata.

Teen-way rule (sab cases):

• $Q<K_{sp}$ ⇒ unsaturated: koi solid nahi, aur dissolve ho sakta hai.
• $Q=K_{sp}$ ⇒ saturated equilibrium: solution exactly on the red line hai. Yahan koi net change nahi — solid aur ions coexist karte hain, equal rates par dissolve aur precipitate hote hue. Kuch naya precipitate nahi hota aur kuch extra dissolve nahi hota; system already rest par hai.
• $Q>K_{sp}$ ⇒ supersaturated: precipitation tab tak chalta hai jab tak $Q$ vapas $K_{sp}$ par nahi aa jaata.

Boundary case $Q=K_{sp}$ exactly woh "just begins" condition hai jise hum 3.3 aur 4.1 mein exploit karte hain.

WHAT: $s$ = mol/L of AgBr jo dissolve hota hai. WHY $s$ dono ions ke liye: ratio $1:1$ hai, to har formula unit ek ${\text{Ag}}^{+}$ aur ek ${\text{Br}}^{-}$ release karta hai. $[{\text{Ag}}^{+}]=s,[{\text{Br}}^{-}]=s$ $K_{sp}=s\cdot s=s^2\Rightarrow s=\sqrt{5.0\times 10^{-13}}=7.07\times 10^{-7}\text{mol/L}$

WHAT — ion concentrations set karo. Har dissolved ${\text{PbCl}}_2$ ek ${\text{Pb}}^{2+}$ aur do ${\text{Cl}}^{-}$ deta hai, to agar $s$ mol/L dissolve hote hain, $[{\text{Pb}}^{2+}]=s$ aur $[{\text{Cl}}^{-}]=2s$. WHY $2s$: balanced equation har formula unit per do chlorides release karta hai, isliye unki concentration salt ki concentration se twice as fast badhti hai. $[{\text{Pb}}^{2+}]=s,[{\text{Cl}}^{-}]=2s$ WHAT — $K_{sp}$ mein substitute karo; WHY: saturation par ion product hai hi $K_{sp}$, to $s$-expressions plug in karna equilibrium condition ko $s$ mein ek equation mein badal deta hai. $K_{sp}=s(2s{)}^2=4s^3$ WHAT — solve karo; WHY cube root: unknown $s^3$ ke roop mein appear hota hai, to cube undo karna $s$ ko isolate karta hai. $s^3=\frac{1.6\times 10^{-5}}{4}=4.0\times 10^{-6}\Rightarrow s=(4.0\times 10^{-6}{)}^{1/3}=1.587\times 10^{-2}\text{mol/L}$

WHAT — stoichiometry ulta padhna; WHY: is baar hume $s$ diya gaya hai aur $K_{sp}$ chahiye, to pehle measured solubility ko ion concentrations mein translate karte hain. ${\text{Mg(OH)}}_2\rightleftharpoons {\text{Mg}}^{2+}+2{\text{OH}}^{-}$ deta hai $[{\text{Mg}}^{2+}]=s$ aur $[{\text{OH}}^{-}]=2s$ (do hydroxides per formula unit, same reason as 2.2). WHAT — $K_{sp}$ banao; WHY: $K_{sp}$ by definition saturation par ion product hai, aur measured $s$ hai saturation point, to hum sirf evaluate karte hain. $K_{sp}=s(2s{)}^2=4s^3=4(1.65\times 10^{-4}{)}^3=1.80\times 10^{-11}$

Notation pehle: naye solubility ko $s^{\prime }$ (padho "$s$-prime") likhte hain, purely pure-water value $s$ of Exercise 2.1 se distinguish karne ke liye — same meaning (mol/L of AgBr jo dissolve hota hai), bas naye common-ion conditions ke under. Prime ek label hai, naya quantity nahi. WHAT: $\text{KBr}$ fully soluble hai, to woh solution mein ${\text{Br}}^{-}$ flood karta hai pehle se, koi AgBr dissolve hone se pehle. Phir: $[{\text{Ag}}^{+}]=s^{\prime },[{\text{Br}}^{-}]=0.10+s^{\prime }$ WHY approximation: AgBr itna kanjoos hai ki $s^{\prime }\ll 0.10$, to ${\text{Br}}^{-}$ jo woh add karta hai already wahan maujood $0.10\text{M}$ par rounding error hai — isliye $[{\text{Br}}^{-}]\approx 0.10$. $K_{sp}=s^{\prime }(0.10)\Rightarrow s^{\prime }=\frac{5.0\times 10^{-13}}{0.10}=5.0\times 10^{-12}\text{mol/L}$ Check: $5.0\times 10^{-12}\ll 0.10$ ✓. Pure water ($7.07\times 10^{-7}$) ke compare mein, solubility $\approx 1.4\times 10^5$ factor se drop hoti hai — common ion isse crush kar deta hai.

Figure 3 is suppression ko ek bar chart par dikhata hai: pure-water solubility common-ion solubility ke upar tower karti hai, shared ${\text{Br}}^{-}$ ki flood se neeche drive hoti hui.

WHAT pehle — dilution. Equal volumes mix karna har concentration ko halve kar deta hai, kyunki har ion ab $100\text{mL}$ ki jagah $200\text{mL}$ mein spread ho jaata hai. WHY: moles conserve hote hain, volume double hota hai, to concentration halve hoti hai. $[{\text{Ag}}^{+}]=\frac{0.010}{2}=5.0\times 10^{-3}\text{M},[{\text{Cl}}^{-}]=5.0\times 10^{-3}\text{M}$ Ab current reading $Q$ compute karo: $Q=[{\text{Ag}}^{+}][{\text{Cl}}^{-}]=(5.0\times 10^{-3}{)}^2=2.5\times 10^{-5}$ $Q=2.5\times 10^{-5}\gg K_{sp}=1.8\times 10^{-10}$ $Q>K_{sp}$ ⇒ haan, AgCl precipitate karta hai.

Figure 4 mixing draw karta hai: do $100\text{mL}$ beakers ek $200\text{mL}$ beaker mein pour hote hain, aur concentration bars halve ho jaate hain — visual reason kyun $Q$ diluted numbers use karna chahiye.

Precipitation exactly us instant shuru hoti hai jab $Q=K_{sp}$ (1.2 se boundary case), to unhe equal set karo aur unknown ion ke liye solve karo. $K_{sp}=[{\text{Mg}}^{2+}][{\text{OH}}^{-}{]}^2\Rightarrow [{\text{OH}}^{-}]=\sqrt{\frac{K_{sp}}{[{\text{Mg}}^{2+}]}}=\sqrt{\frac{1.8\times 10^{-11}}{0.010}}$ $[{\text{OH}}^{-}]=\sqrt{1.8\times 10^{-9}}=4.24\times 10^{-5}\text{M}$ Phir $\text{pOH}=-\log (4.24\times 10^{-5})=4.37$, to (dekho pH and pOH Calculations): $\text{pH}=14-4.37=9.63$

WHAT — har salt ka trigger nikalo; WHY: precipitation us pal shuru hoti hai jab $Q=K_{sp}$ (from 1.2), to har salt ke liye hum $[{\text{Ag}}^{+}]$ solve karte hain jo pehle apni red line reach karta hai. Jis ki red line lower $[{\text{Ag}}^{+}]$ par reach hoti hai woh pehle trip karta hai, kyunki hum $[{\text{Ag}}^{+}]$ gradually zero se badha rahe hain.

AgCl ke liye (1:1): $[{\text{Ag}}^{+}{]}_{\text{needed}}=\frac{K_{sp}}{[{\text{Cl}}^{-}]}=\frac{1.8\times 10^{-10}}{0.010}=1.8\times 10^{-8}\text{M}$ ${\text{Ag}}_2{\text{CrO}}_4$ ke liye (2:1) — WHY square: balanced dissolution do ${\text{Ag}}^{+}$ release karta hai per formula unit, to $[{\text{Ag}}^{+}]$ $K_{sp}$ mein squared enter karta hai, aur undo karne ke liye square root chahiye: $K_{sp}=[{\text{Ag}}^{+}{]}^2[{\text{CrO}}_4^{2-}]\Rightarrow [{\text{Ag}}^{+}{]}_{\text{needed}}=\sqrt{\frac{1.1\times 10^{-12}}{0.010}}=\sqrt{1.1\times 10^{-10}}=1.049\times 10^{-5}\text{M}$ AgCl ko sirf $1.8\times 10^{-8}\text{M}$ chahiye vs chromate ke liye $1.049\times 10^{-5}\text{M}$ — AgCl pehle precipitate karta hai (~580× jaldi). Yeh Mohr titration ka basis hai Qualitative Inorganic Analysis mein.

WHAT — $[{\text{Ag}}^{+}]$ ko doosre salt ke trigger par fix karo; WHY: chromate exactly us pal shuru hota hai jab $[{\text{Ag}}^{+}]=1.049\times 10^{-5}\text{M}$ (from 4.1), aur us same instant solution mein bacha chloride AgCl ke apne $K_{sp}$ ko obey karta hai — kyunki AgCl already saturated hai aur apni red line par baitha hai. To hum leftover chloride seedha us equation se read karte hain. $[{\text{Cl}}^{-}{]}_{\text{left}}=\frac{K_{sp}(\text{AgCl})}{[{\text{Ag}}^{+}]}=\frac{1.8\times 10^{-10}}{1.049\times 10^{-5}}=1.716\times 10^{-5}\text{M}$ WHAT — fraction mein convert karo; WHY: leftover ko original $0.010\text{M}$ se divide karna hume woh share batata hai chloride ka jo abhi tak precipitate nahi hua — yeh measure hai ki cut kitna clean hai. $\frac{1.716\times 10^{-5}}{0.010}=1.716\times 10^{-3}=0.172\%$ $99.8\%$ se zyada chloride already AgCl ke roop mein neeche aa chuka hai pehle doosra salt shuru ho — ek clean separation.

Step 1 — dilute karo; WHY: equal volumes mix hone ka matlab hai har species ab $50\text{mL}$ ki jagah $100\text{mL}$ occupy karti hai, to har concentration halve ho jaati hai (moles conserved, volume doubled). $[{\text{Pb}}^{2+}{]}_0=\frac{0.20}{2}=0.10\text{M},[{\text{Cl}}^{-}{]}_0=\frac{0.20}{2}=0.10\text{M}$ Step 2 — $Q$ se test karo; WHY: hum current ion product compute karte hain dekhne ke liye ki kya yeh red line exceed karta hai (rule 1.2). $Q=[{\text{Pb}}^{2+}][{\text{Cl}}^{-}{]}^2=(0.10)(0.10{)}^2=1.0\times 10^{-3}\gg 1.6\times 10^{-5}=K_{sp}$ $Q>K_{sp}$ ⇒ precipitation occurs. (a) shown.

Step 3 — completion tak react karo; WHY limiting ion: har ${\text{Pb}}^{2+}$ do ${\text{Cl}}^{-}$ consume karta hai, to $0.10\text{M}$ chloride sirf $0.05\text{M}$ lead ko down kar sakta hai — chloride pehle khatam hota hai, isse limiting reagent banata hai. ${\text{Pb}}^{2+}+2{\text{Cl}}^{-}\to {\text{PbCl}}_2$ ke baad: $[{\text{Pb}}^{2+}{]}_{\text{excess}}=0.10-0.05=0.05\text{M}$ Step 4 — $[{\text{Cl}}^{-}]\approx 0$ justify karo; WHY it is safe: thodi si ${\text{PbCl}}_2$ jo redissolve hoti hai woh large lead excess ke against equilibrium se capped hai. $[{\text{Pb}}^{2+}]\approx 0.05\text{M}$ ke saath redissolved chloride hai $[{\text{Cl}}^{-}]=\sqrt{\frac{K_{sp}}{[{\text{Pb}}^{2+}]}}=\sqrt{\frac{1.6\times 10^{-5}}{0.05}}=\sqrt{3.2\times 10^{-4}}=1.79\times 10^{-2}\text{M}$ Yeh $1.79\times 10^{-2}\text{M}$ us $\approx 0.05\text{M}$ chloride se kaafi chhoti hai jo solid ke roop mein precipitate hui, to post-reaction chloride ko "almost gone" treat karna lead balance ko thoda bhi disturb nahi karta — excess lead $\approx 0.05\text{M}$ rehta hai. (b) answer: $\boxed{[{\text{Pb}}^{2+}]\approx 0.05\text{M}}$

Lower edge — Fe $10^{-6}$ M tak remove; WHY: hum poochh rahe hain ki kaun sa $[{\text{OH}}^{-}]$ iron ko target $10^{-6}\text{M}$ tak neeche dhakelta hai, Fe ke apne $K_{sp}$ use karke (teen hydroxides ⇒ cube): $[{\text{OH}}^{-}{]}^3=\frac{2.6\times 10^{-39}}{10^{-6}}=2.6\times 10^{-33}\Rightarrow [{\text{OH}}^{-}]=(2.6\times 10^{-33}{)}^{1/3}=1.375\times 10^{-11}\text{M}$ $\text{pOH}=-\log (1.375\times 10^{-11})=10.86\Rightarrow \text{pH}=14-10.86=3.14$ To pH 3.14 se upar iron essentially gone hai.

Upper edge — Mg shuru nahi hona chahiye; WHY: hum woh $[{\text{OH}}^{-}]$ nikalte hain jis par Mg pehli baar apni red line reach karta hai ("just begins" condition from 1.2); usse neeche rehna Mg ko dissolved rakhta hai. $[{\text{OH}}^{-}]=\sqrt{\frac{K_{sp}}{[{\text{Mg}}^{2+}]}}=\sqrt{\frac{1.8\times 10^{-11}}{0.010}}=\sqrt{1.8\times 10^{-9}}=4.24\times 10^{-5}\text{M}$ $\text{pOH}=-\log (4.24\times 10^{-5})=4.37\Rightarrow \text{pH}=14-4.37=9.63$ To Mg pH 9.63 se neeche dissolved rehta hai.

Window: pH ko $\approx 3.1$ aur $\approx 9.6$ ke beech rakho (e.g. pH 5 ke paas ek acetate buffer) taaki Fe(OH)₃ cleanly drop ho jabki Mg²⁺ solution mein rahe. Yeh precisely Group-III hydroxide separation hai Qualitative Inorganic Analysis mein, aur thermodynamic direction $\Delta G<0$ se fix hai (dekho Gibs Free Energy and Spontaneity).

WHY $K_{sp}$ akela kaafi nahi hai; WHAT naya hai: naive picture kehti hai "zyada ${\text{OH}}^{-}$ ⇒ zyada precipitate" (common-ion, Le Chatelier). Lekin aluminium ka ek doosra channel hai — solid excess base ke saath react kar sakta hai ek soluble complex banane ke liye. To total dissolved Al do routes ka sum hai.

Route 1 — free ion via $K_{sp}$: $[{\text{Al}}^{3+}]=\frac{K_{sp}}{[{\text{OH}}^{-}{]}^3}=\frac{3.0\times 10^{-34}}{(1.0{)}^3}=3.0\times 10^{-34}\text{M}$ Vanishingly small — common-ion route kehta hai aluminium "gone" hai.

Route 2 — complex via $K_c$; WHY: solid jo ${\text{Al(OH)}}_4^{-}$ ke roop mein dissolve hota hai woh apne equilibrium ko obey karta hai, aur pure solid ki activity $1$ ke saath: $K_c=\frac{[{\text{Al(OH)}}_4^{-}]}{[{\text{OH}}^{-}]}\Rightarrow [{\text{Al(OH)}}_4^{-}]=K_c[{\text{OH}}^{-}]=40\times 1.0=40\text{M (formally)}$ Ise yun padho "complex route enormously favoured hai" — practically sab solid dissolve ho jaata hai $40\text{M}$ pahunchne se pehle; number batata hai ki effectively koi solid nahi bacha.

(a) Total dissolved aluminium $\approx 3.0\times 10^{-34}+40\approx 40\text{M}$ formally — entirely complex se dominated. (b) ${\text{Al(OH)}}_4^{-}$ term free-ion term ko ~35 orders of magnitude se outweigh karta hai. Lesson: kabhi bhi ek amphoteric hydroxide precipitate (${\text{Al}}^{3+}$, ${\text{Zn}}^{2+}$, ${\text{Pb}}^{2+}$, ${\text{Cr}}^{3+}$) ko excess strong base se wash mat karo — wahi ion jo tumne precipitate karne ki koshish ki thi complex ke roop mein re-dissolve ho jaata hai, aur $K_{sp}$ reasoning silently fail ho jaati hai kyunki ek doosra equilibrium khul gaya hai. Yeh Group-III analysis ka amphoteric trap hai Qualitative Inorganic Analysis mein.

WHY ideal picture fail hoti hai; WHAT badalta hai: ${\text{KNO}}_3$ koi ion share nahi karta, to common-ion reasoning predict karta hai koi effect nahi. Lekin true $K_{sp}$ activities $a=\gamma [\text{ion}]$ se bana hai, concentrations se nahi. Real constant hai $K_{sp}=a_{\text{Ag}}{a}_{\text{Cl}}={\gamma }_{\pm }^2[{\text{Ag}}^{+}][{\text{Cl}}^{-}]$ Concentration product ke liye solve karte hue (jo actually dissolve hota hai): $[{\text{Ag}}^{+}][{\text{Cl}}^{-}]=\frac{K_{sp}}{{\gamma }_{\pm }^2}=\frac{1.8\times 10^{-10}}{(0.75{)}^2}=3.2\times 10^{-10}$ $[{\text{Ag}}^{+}]=[{\text{Cl}}^{-}]=s$ ke saath: $s=\sqrt{3.2\times 10^{-10}}=1.789\times 10^{-5}\text{M}$ Direction aur lesson: $s$ $1.34\times 10^{-5}$ se badh gaya $1.79\times 10^{-5}\text{M}$ tak — lagbhag $33\%$ zyada soluble, even though humne ek salt add kiya jisme koi common ion nahi tha. WHY badhta hai: ${\text{K}}^{+}/{\text{NO}}_3^{-}$ ka ionic atmosphere har ${\text{Ag}}^{+}$ aur ${\text{Cl}}^{-}$ ko shield karta hai, unki effective (activity) concentration kam karta hai; true activity product ko $K_{sp}$ par rakhne ke liye zyada solid dissolve hona padta hai. Yeh salting-in / inert-salt effect hai — page ke top par kholi gayi activity caveat ka concrete pay-off. Zyada Ionic Equilibria in Solutions mein.