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A saturated solution of o- nitrophenol h...

A saturated solution of o- nitrophenol has a pH equal to 4.53 then its solubility in water is `(pK_a=7.23) `

A

` 2.085 g//"lit"`

B

` 20. 85 g //"lit"`

C

` 10. 425g//"lit"`

D

` 1.0425 g//"lit"`

Text Solution

Verified by Experts

The correct Answer is:
A
` O- " nitrophenol" _(s) hArr O- " nitro phenol "_((aq) ) `
` O- "nitrophenol"_((aq)) hArr O- " nitro phenol " + H^(+) `
` S(1-alpha )" " S alpha " " S alpha `
` [H^(+) ]=S alpha =sqrt( SKa) `
` -log [H^(+) ] =(-log S -log Ka)/( 2) `
` pH =(-log S - log Ka )/( 2) `
` pH =(- log S +pKa)/( 2) `
` - log S = 2xx 4.53 -7.23 =1.83`
` rArr S = 1 .48 xx 10 ^(-2) "mol" //"lit" `
` M. wt = 139 rArr S =2.06 g//lit `
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A solution which remains in equilibrium with undissolved solute is said to be saturated. The concentration of a saturated solution at a given temperature is called solubility. The product of concentration of ions in a saturated solution of an electrolyte at a given temperature, is called solubility product (K_(sp)) . For the electrolyte, A_(x),B_(y),:A_(x),B_(y(s)) rarr xA^(y+)+ y^(Bx-) , with solubility S, the solubility product (K_(sp)) =x^(x)xxy^(y) xx s^(x+y) . While calculating the solubility of a sparingly soluble salt in the presence of some strong electrolyte containing a common ion, the common ion concentration is practically equal to that of strong electrolyte. If in a solution, the ionic product of an clectrolyte exceeds its K_(sp) , value at a particular temperature, then precipitation occurs. The solubility of PbSO_(4) , in water is 0.303 g/l at 25^(@) C, its solubility product at that temperature is

A solution which remains in equilibrium with undissolved solute is said to be saturated. The concentration of a saturated solution at a given temperature is called solubility. The product of concentration of ions in a saturated solution of an electrolyte at a given temperature, is called solubility product (K_(sp)) . For the electrolyte, A_(x),B_(y),:A_(x),B_(y(s)) rarr xA^(y+)+ y^(Bx-) , with solubility S, the solubility product (K_(sp)) =x^(x)xxy^(y) xx s^(x+y) . While calculating the solubility of a sparingly soluble salt in the presence of some strong electrolyte containing a common ion, the common ion concentration is practically equal to that of strong electrolyte. If in a solution, the ionic product of an clectrolyte exceeds its K_(sp) , value at a particular temperature, then precipitation occurs. The solubility of BaSO_(4) , in 0.1 M BaCl_(2) , solution is (K_(sp) , of BaSO_(4), = 1.5 xx 10^(-9))

Knowledge Check

  • The solublity of I_2 in KI solution is more than its solubility in pure water because

    A
    `I_2` dissociates in water
    B
    `I_2` does not react with water
    C
    `I_2` forms soluble complex `KI_3` with KI
    D
    None of these

  • Sparingly soluble salts maintains their solubility product value in their saturated solutions irrespective of the sources of the ions . what is the molar solubility of AgNO_3 in a 0.1 M H_2S solution buffered at pH= 2 (K_1 and K_2 " of " H_2S " are " 10^(4) and 10^(-8) respectively) (K_(sp) " of " Ag_2S = 4xx 10^(-13) ) (Note : No Ag_2S precipitate should be formed)

    A
    ` 0.01 M`
    B
    ` 0. 02 M`
    C
    ` 0. 03 M`
    D
    ` 0. 04 M`

  • pH of Ba(OH)_2 solution is 12. Its solubility product is:

    A
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    B
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    C
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    D
    ` 5xx 10 ^(-7) M^(3)`

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