If 0.015 M NaCl shows conductivity of 12,000 units, based on \"total ion concent

Question

If 0.015 M NaCl shows conductivity of 12,000 units, based on "total ion concentration" what conductivity should be observed for 0.015 M K2SO4? (WITH EXPLANATION)

Explanation / Answer

If conductivity is purely based on total ion concentration this is straightforward. NaCl -> Na^+ + Cl^- (two ions in solution assuming complete dissociation) and has conductivity of 12,000 units. For the other solution at the same concentration and conditions: K2SO4 -> 2K^+ + SO4^2- (three ions in solution, one of which can transfer 2 electrons, therefore with capacity to carry double the amount of charge of the NaCl). So the current carrying capacity of the 0.015M K2SO4 is 12,000 * 2 = 24,000 units

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