Exercises — Common ion effect

(a) ${\text{CaF}}_2\rightleftharpoons {\text{Ca}}^{2+}+2{\text{F}}^{-}$. NaF extra ${\text{F}}^{-}$ supply karta hai — common ion hai ${\text{F}}^{-}$. Equilibrium left shift karta hai, isliye solubility down jaati hai.

(b) $\text{AgCl}\rightleftharpoons {\text{Ag}}^{+}+{\text{Cl}}^{-}$. ${\text{KNO}}_3$ se ${\text{K}}^{+}$ aur ${\text{NO}}_3^{-}$ milte hain — koi bhi AgCl equilibrium mein appear nahi karta. Koi common ion nahi, isliye pehli approximation mein solubility unchanged rehti hai.

(c) ${\text{BaSO}}_4\rightleftharpoons {\text{Ba}}^{2+}+{\text{SO}}_4^{2-}$. ${\text{Na}}_2{\text{SO}}_4$ se ${\text{SO}}_4^{2-}$ milta hai — common ion hai ${\text{SO}}_4^{2-}$. Solubility down jaati hai.

Dissolving: ${\text{Ag}}_2{\text{CrO}}_4\text{(s)}\rightleftharpoons 2{\text{Ag}}^{+}+{\text{CrO}}_4^{2-}$.

$K_{sp}=[{\text{Ag}}^{+}{]}^2[{\text{CrO}}_4^{2-}]$

Pure water mein, agar $s$ mol salt per litre dissolve hoti hai toh yeh $2s$ of ${\text{Ag}}^{+}$ aur $s$ of ${\text{CrO}}_4^{2-}$ banata hai:

$K_{sp}=(2s{)}^2(s)=4s^3⟹s={\left(\frac{K_{sp}}{4}\right)}^{1/3}$

Square ke andar $2$ kyun hai? Coefficient $2$ concentration multiply karta hai ($[{\text{Ag}}^{+}]=2s$) aur exponent $2$ hai kyunki ion product mein do baar appear karta hai. Dono effects alag hain; inhe merge mat karo.

${\text{CaF}}_2\rightleftharpoons {\text{Ca}}^{2+}+2{\text{F}}^{-}$, toh $[{\text{Ca}}^{2+}]=s$, $[{\text{F}}^{-}]=2s$.

$K_{sp}=s(2s{)}^2=4s^3$ $s={\left(\frac{3.9\times 10^{-11}}{4}\right)}^{1/3}={\left(9.75\times 10^{-12}\right)}^{1/3}=2.14\times 10^{-4}\text{M}$

NaF se $[{\text{F}}^{-}]=0.10\text{M}$ milta hai ${\text{CaF}}_2$ ke react karne se pehle. Nayi solubility $=s^{\prime }$ maano. $[{\text{Ca}}^{2+}]=s^{\prime },[{\text{F}}^{-}]=0.10+2s^{\prime }$ Expect karo $s^{\prime }$ bahut chhota hoga, isliye $0.10+2s^{\prime }\approx 0.10$: $K_{sp}=s^{\prime }(0.10{)}^2⟹s^{\prime }=\frac{3.9\times 10^{-11}}{0.01}=3.9\times 10^{-9}\text{M}$ Check: $2s^{\prime }=7.8\times 10^{-9}\ll 0.10$ ✓ (well under $5\%$). Solubility $2.14\times 10^{-4}$ se gir ke $3.9\times 10^{-9}$ ho gayi — lagbhag 55 000× ki drop.

${\text{PbI}}_2\rightleftharpoons {\text{Pb}}^{2+}+2{\text{I}}^{-}$. KI se $[{\text{I}}^{-}]=0.010$ milta hai. $[{\text{Pb}}^{2+}]=s^{\prime },[{\text{I}}^{-}]=0.010+2s^{\prime }$ Shortcut (assume $0.010+2s^{\prime }\approx 0.010$): $K_{sp}=s^{\prime }(0.010{)}^2⟹s^{\prime }=\frac{7.1\times 10^{-9}}{1.0\times 10^{-4}}=7.1\times 10^{-5}\text{M}$ Validity check: dissolution se extra iodide hai $2s^{\prime }=1.42\times 10^{-4}$. $0.010$ se compare karo: yeh $1.42\%$ hai. $5\%$ se kam, isliye approximation yahan valid hai. ✓

Parent ke PbCl₂ example se contrast karo, jahan common ion cation tha aur dissolved salt ne doosre ion ka bahut zyada bana diya — wahan approximation fail ho gayi thi. Lesson yeh hai: $s^{\prime }$ ka $c$ ke relative size decide karta hai, problem ki wording nahi.

${\text{Pb}}^{2+}$ add karo ($c=0.010$): $[{\text{Pb}}^{2+}]\approx 0.010$, $[{\text{I}}^{-}]=2s^{\prime }$. $K_{sp}=0.010(2s^{\prime }{)}^2⟹(s^{\prime }{)}^2=\frac{7.1\times 10^{-9}}{0.04}=1.775\times 10^{-7}$ $s^{\prime }=4.21\times 10^{-4}\text{M}$ ${\text{I}}^{-}$ add karo ($c=0.010$): 3.1 se, $s^{\prime }=7.1\times 10^{-5}\text{M}$.

Conclusion: ${\text{I}}^{-}$ add karna zyada suppress karta hai ($7.1\times 10^{-5}$ vs $4.21\times 10^{-4}$, roughly $6\times$ zyada strong).

Kyun? ${\text{I}}^{-}$ $K_{sp}$ mein squared appear karta hai. Ek squared term ko force up karna equilibrium ko kahin zyada cost karta hai, isliye salt ko $K_{sp}$ constant rakhne ke liye kam dissolve karna padta hai. Higher exponent wale ion par attack karo.

Har precipitate start karne ke liye needed ${\text{Ag}}^{+}$ (reaction quotient $=K_{sp}$ set karo):

• AgCl: $[{\text{Ag}}^{+}]=\frac{K_{sp}}{[{\text{Cl}}^{-}]}=\frac{1.8\times 10^{-10}}{0.020}=9.0\times 10^{-9}\text{M}$
• ${\text{Ag}}_2{\text{CrO}}_4$: $[{\text{Ag}}^{+}]=\sqrt{\frac{K_{sp}}{[{\text{CrO}}_4^{2-}]}}=\sqrt{\frac{1.1\times 10^{-12}}{0.020}}=\sqrt{5.5\times 10^{-11}}=7.42\times 10^{-6}\text{M}$

AgCl ko bahut kam ${\text{Ag}}^{+}$ chahiye, isliye AgCl pehle precipitate karta hai.

${\text{Ag}}_2{\text{CrO}}_4$ just start hone par $[{\text{Cl}}^{-}]$ (yaani jab $[{\text{Ag}}^{+}]=7.42\times 10^{-6}$): $[{\text{Cl}}^{-}]=\frac{K_{sp}(\text{AgCl})}{[{\text{Ag}}^{+}]}=\frac{1.8\times 10^{-10}}{7.42\times 10^{-6}}=2.43\times 10^{-5}\text{M}$ Toh jab tak chromate girna shuru hota hai, chloride $2.43\times 10^{-5}\text{M}$ tak aa jaata hai — chloride ka $99.8\%$ se zyada already remove ho chuka hota hai. Clean separation.

${\text{CH}}_3\text{COOH}\rightleftharpoons {\text{H}}^{+}+{\text{CH}}_3{\text{COO}}^{-}$. $K_a=\frac{[{\text{H}}^{+}][{\text{CH}}_3{\text{COO}}^{-}]}{[{\text{CH}}_3\text{COOH}]}$ Common ion ke saath (acetate $\approx 0.10$, acid $\approx 0.10$): $[{\text{H}}^{+}]=K_a\frac{[{\text{CH}}_3\text{COOH}]}{[{\text{CH}}_3{\text{COO}}^{-}]}=1.8\times 10^{-5}\cdot \frac{0.10}{0.10}=1.8\times 10^{-5}\text{M}$ Akele acid ($0.10\text{M}$): $[{\text{H}}^{+}]=\sqrt{K_a\cdot 0.10}=\sqrt{1.8\times 10^{-6}}=1.34\times 10^{-3}\text{M}$. Common ion add karne se $[{\text{H}}^{+}]$ $1.34\times 10^{-3}$ se $1.8\times 10^{-5}$ tak aa gaya — ionisation ka 74× suppression. Same physics jaise solubility suppression, weak acid par apply kiya.

${\text{PbCl}}_2\rightleftharpoons {\text{Pb}}^{2+}+2{\text{Cl}}^{-}$. NaCl se $[{\text{Cl}}^{-}]=0.10$ milta hai. $[{\text{Pb}}^{2+}]=s^{\prime },[{\text{Cl}}^{-}]=0.10+2s^{\prime }$ Shortcut ($0.10+2s^{\prime }\approx 0.10$): $s^{\prime }=\frac{K_{sp}}{(0.10{)}^2}=\frac{1.7\times 10^{-5}}{0.010}=1.7\times 10^{-3}\text{M}$ Check: dissolution $2s^{\prime }=3.4\times 10^{-3}$ add karta hai, jo $0.10$ ka $3.4\%$ hai. $5\%$ se kam, isliye shortcut valid hai ✓. $s^{\prime }\approx 1.7\times 10^{-3}\text{M}$.

Note karo yeh parent ke example se kitna alag behave karta hai jahan ${\text{Pb}}^{2+}$ common ion tha: wahan chloride (squared, dissolution se) badh ke approximation tod deti thi. Kaun sa ion add karte ho yeh sab kuch change kar deta hai — squared ion par attack karna (jaise yahan) hardest suppress karta hai aur $s^{\prime }$ chhota rakhta hai.

${\text{Mg(OH)}}_2\rightleftharpoons {\text{Mg}}^{2+}+2{\text{OH}}^{-}$, isliye $K_{sp}=[{\text{Mg}}^{2+}][{\text{OH}}^{-}{]}^2$. $\text{pH}=9.00$ par: $[{\text{H}}^{+}]=10^{-9}$, isliye $[{\text{OH}}^{-}]=\frac{10^{-14}}{10^{-9}}=10^{-5}\text{M}$ (buffer se fixed). $[{\text{Mg}}^{2+}]=\frac{K_{sp}}{[{\text{OH}}^{-}{]}^2}=\frac{5.6\times 10^{-12}}{(10^{-5}{)}^2}=\frac{5.6\times 10^{-12}}{10^{-10}}=5.6\times 10^{-2}\text{M}$ Kyunki har dissolved ${\text{Mg(OH)}}_2$ ek ${\text{Mg}}^{2+}$ deta hai, solubility $=[{\text{Mg}}^{2+}]=5.6\times 10^{-2}\text{M}$.

Insight: yahan zyada ${\text{OH}}^{-}$ (more basic) solubility ghata dega (common ion), aur buffer hume $[{\text{OH}}^{-}]$ ko kisi bhi value par dial karne deta hai. $\text{pH}=9$ par hydroxide itna kam hai ki ${\text{Mg(OH)}}_2$ kaafi soluble hai. Exactly yahi hota hai jab ${\text{Mg(OH)}}_2$ acid mein dissolve hota hai: acid ${\text{OH}}^{-}$ destroy karta hai, dissolution force karta hai — anti-common-ion effect.