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The solubility of salt of weak acid MX (e.g. phosphoric) is increase at tower pH explain with equation. |
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Answer» Solution :The solubility of salt of weak acid like `Na_3PO_4` increases at lower pH because as lower pH concentration of anion `X^-` decreases due to protonation so `X^-` decrease and solubility of MX increases. `MX HARR M_((aq))^(+) + X_((aq))^(-)`…(Eq. -i) `K_(sp)=[M^+][X^-]`...(Eq.-ii) MX is the salt of weak acid (HX) ionization of weak acid is as follows . `HX_((aq)) hArr H_((aq))^(+) + X_((aq))^(-)` ...(Eq.-iii) (Eq.-iii) ionization constant for weak acid `K_a=([H_((aq))^+][X_(aq)^-])/(HX_((aq))]`...(Eq.-iv) `THEREFORE K_a/([H^+])=([X^-])/([HX])` `([H^+])/K_a +1 = ([HX])/([X^-])+1` `because ([H^+]+K_a)/K_a=([HX]+[X^-])/([X^-])` `K_a/([H^+]+K_a)=([X^-])/([HX]+[X^-])=F` ....(Eq.-v) So `[H^+]` increase pH decreases and with of .f. decrease. SUPPOSE , for given salt MX at given pH .S. is molar . `K_(sp)=[M^+][X^-]=S-(f S) =S^2(f)` `therefore S=sqrt(K_(sp)/f)` ...(Eq.-vi) As per (Eq.-v) `1/f=(([H^+]+K_a)/K_a)`....(Eq.-vii) `S=((K_(sp){(H^+)+(K_a)})/K_a)^(1/2)` ....(Eq.-viii) As per this EQUATION `S prop [H^+] pH prop 1/([H^+])` |
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