a) What is the K sp for the lithium tetraborate if50 mL of a saturated solution

Question

a) What is the Ksp for the lithium tetraborate if50 mL of a saturated solution of lithium tetraborate requires9.11mL of 0.511 M HCl to titrate the tetraborate? b) Based on the Ksp you calculated in the prviousquestion, what id G for the dissolution of lithiumtetraborate? a) What is the Ksp for the lithium tetraborate if50 mL of a saturated solution of lithium tetraborate requires9.11mL of 0.511 M HCl to titrate the tetraborate? b) Based on the Ksp you calculated in the prviousquestion, what id G for the dissolution of lithiumtetraborate?

Explanation / Answer

The stoichiometric equation of the titrationof
          B4O7-2 (aq) + 2 HCl (aq)---------> 2LiCl(aq) +H2B4O7 Thus every 1mole of HClconsumes 1/2 moles of tetraborate ion. Thus tertaborate ion that is consumed by thegiven Hcl solution                                    = (0.511M*0.00911 L)*1 mole tertaborate ion/2 moleHCl)                                    = 0.00232 mole Thus concentration of Oxalate in solution = 0.00232mole/0.05L                                                               = 0.04655 mole/L This is the molar solubilituy of each ion in theLi2B4O7 Thus Ksp =[Li+]2[B4O7-2]                 = (4s3                  =4*(0.04655)3                 = 4.034*10-4 b) We know that G = -RTlnKsp                                   = -8.314 * 298*ln(4.034*10-4 )                                   = 19363.66 J                                   = 19.363 kJ b) We know that G = -RTlnKsp                                   = -8.314 * 298*ln(4.034*10-4 )                                   = 19363.66 J                                   = 19.363 kJ

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