Given that the solubility product of radium sulphate ( RaSO 4 ) is 4 × 10 - 11 . Calculate the solubility in pure water.

The solubility equilibrium for RaSO 4 may be represented as: RaSO 4 ⇌ Ra 2 + + SO 4 2 - Let the solubility of RaSO 4 be s moles / litre . Then the solution will contain s moles of Ra 2 + ions and s moles of SO 4 2 - ions respectively per litre. Hence, the solubility product, K sp of RaSO 4 , would be given by the expression, K sp = [ Ra 2 + ] [ SO 4 2 - ] = ( s ) × ( s ) = s 2 But the value of K sp = 4 × 10 - 11 (Given) ∴ 4 × 10 - 11 = s 2 Or s = 6 . 66 × 10 - 6 mol L - 1 .

Given that the solubility product of radium sulphate ( RaSO 4 ) is 4 × 10 - 11 . Calculate the solubility in pure water.

The solubility equilibrium for RaSO 4 may be represented as: RaSO 4 ⇌ Ra 2 + + SO 4 2 - Let the solubility of RaSO 4 be s moles / litre . Then the solution will contain s moles of Ra 2 + ions and s moles of SO 4 2 - ions respectively per litre. Hence, the solubility product, K sp of RaSO 4 , would be given by the expression, K sp = [ Ra 2 + ] [ SO 4 2 - ] = ( s ) × ( s ) = s 2 But the value of K sp = 4 × 10 - 11 (Given) ∴ 4 × 10 - 11 = s 2 Or s = 6 . 66 × 10 - 6 mol L - 1 .

Given that the solubility product of radium sulphate ( RaSO 4 ) is 4 × 10 - 11 . Calculate the solubility in 0 . 10 M Na 2 SO 4 .

Suppose the solubility of RaSO 4 in 0 . 10 M Na 2 SO 4 be s mol L - 1 . RaSO 4 in the solution dissociates as : RaSO 4 ⇌ Ra 2 + + SO 4 2 - Thus, s mol L - 1 of RaSO 4 in the solution gives s mol L - 1 of Ra 2 + ion and s mol L - 1 of SO 4 2 - ions. But, SO 4 2 - ions are also produced from Na 2 SO 4 . As Na 2 SO 4 dissociates completely, SO 4 2 - from Na 2 SO 4 = 0 . 10 M . Hence, [ Ra 2 + ] = s mol L - 1 And, [ SO 4 2 - ] = s + 0 . 1 mol L - 1 ≃ 0 . 1 mol L - 1 ( ∵ s is very small as compared to 0 . 1 ) K sp of RaSO 4 = [ Ra 2 + ] [ SO 4 2 - ] ∴ s ( 0 . 1 ) = 4 × 10 - 11 Or s = 4 × 10 - 10 .

If 20 mL of 2 × 10 - 5 BaCl 2 solution is mixed with 20 mL of 1 × 10 - 5 M Na 2 SO 4 solution, will a precipitate form? ( K sp for BaSO 4 is 1 . 0 × 10 - 10 )

Given: 20 mL of 2 × 10 - 5 M BaCl 2 Mixed with 200 mL of 1 × 10 - 5 M Na 2 SO 4 K sp BaSO 4 = 1 . 0 × 10 - 10 After mixing, the total volume of the solution = ( 20 + 20 ) mL = 40 mL Now, Ba 2 + ions will be formed corresponding to 20 mL of 2 × 10 - 5 M BaCl 2 ∴ Ba 2 + = 20 40 × 2 × 10 - 5 = 10 - 5 M Also, SO 4 2 - ions will be formed corresponding to 20 mL of 1 × 10 - 5 M Na 2 SO 4 ∴ SO 4 2 - = 20 40 × 10 - 5 = 5 × 10 - 6 M Ionic product of BaSO 4 = Ba 2 + SO 4 2 - = 10 - 5 × 5 × 10 - 6 = 5 × 10 - 11 ∵ Ionic Product < K sp , no precipitate will be formed.

The ionization constant of formic acid is 1 . 8 × 10 - 4 . Calculate the ratio of sodium formate and formic acid in a buffer of pH 4 . 25 .

Given: K a for HCOOH = 1 . 8 × 10 - 4 pH of buffer = 4 . 25 To find: [ HCOONa ] [ HCOOH ] So, basically we have to find [ Salt ] [ Acid ] . According to Henderson-Hasselbalch's Equation, pH = pK a + log [ Salt ] [ Acid ] ∴ 4 . 25 = - log 1 . 8 × 10 - 4 + log [ Salt ] [ Acid ] = 3 . 74 + log [ Salt ] [ Acid ] Hence, log [ Salt ] [ Acid ] = 4 . 25 - 3 . 74 Or [ Salt ] [ Acid ] = antilog ( 0 . 51 ) = 3 . 24