(i)Define solubility product.
(ii) What is the pH of an aqueous solution obtained by mixing 6 gram of acetic acid and 8.2 gram of sodium acetate and making the volume equal to 500 ml. (Given: `K_(a)` for acetic acid is `1.8xx10^(-5)`)

Explanation

(i) Solubility product : It is defined as the product of the molar concentration of the constituent ions, each raised to the power of its stoichiometric co-efficient in a balanced equilibrium equation.
`X_(m)Y_(n(s)) overset(H_(2)O)hArr mX^("(aq)") ^(n+)+nY_((aq))^(m-)`
`K_(sp)=[X^(n+)]^(m)[Y^(m-)]^(n)`
(ii) According to Henderson - Hessalbalch equation,
`pH=pK_(a)+log(["salt"])/(["acid"])`, Given that `K_(a)=1.8xx10^(-5)`
`therefore pK_(a)=-"log "K_(a)= -log(1.8xx10^(-5))=4.74`
[Salt] `=("Number of moles of sodium acetate")/("Volume of the solution (litre)")`
Number of moles of sodium acetate `=("mass of sodium acetate")/("molar mass of sodium acetate")=(8.2)/(8.2)=0.1`
`therefore` [Salt] `=("0.1 mole")/(1//2"Litre")=0.2M`
[acid] `=(("mass of "CH_(3)COOH//"molar mass of" CH_(3)COOH))/("Volume of solution in litre")=((6//60))/(1//2)=0.2M`
`therefore pH=4.74+"log" ((0.2))/((0.2))=4.74+log 1 rArr pH=4.74+0=4.74`