A water initially contains 50mg/L of Mg2+. The pH of the water is increased until the...

A water initially contains 50mg/L of Mg2+. The pH of the water is increased until the concentration of hydroxide ion is 0.0005 M. Calculate the concentration (mg/L) of magnesium ion in this water at this pH. Assume that the temperature of the solution is 25 degrees C.

Expert Solution

Given the concentration of hydroxide ion,[OH-] = 0.0005 M

Mg(OH)₂ Mg²⁺ + 2OH^-

2 moles of OH- produced from 1 mole of Mg(OH)₂

So [Mg(OH)₂] = [OH^-]/2

= 0.0005 / 2

= 0.00025 M

We know that Molarity, M = ( mass/molar mass) / volume in L

0.00025 M = ( mass/24.3(g/mol)) / 1L

So mass of Magnesium ion is , m = 6.075x10^-3 g/L

= 6.075 mg/L

queen_honey_blossom answered 2 years ago

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