An aqueous solution of potassium hydroxide (KOH) is fed at a rate of 875 kg/h to an evaporative crystallizer operating at 10°C, producing crystals of [tex]$KOH \cdot 2H_2O$[/tex]. A 5 g aliquot of the feed solution is titrated to neutrality with 22.4 mL of 0.85 molar [tex]$H_2SO_4$[/tex]. The solubility of KOH at 10°C is 103 kg KOH/100 kg [tex]$H_2O$[/tex]. At what rate must water be evaporated to crystallize 60% of the KOH in the feed?

The molar mass of water is 18.0 g/mol, so the mass of water that needs to be evaporated is 18.0 g/mol x 1858.62 mol = 33455.16 g and therefore, the rate of water evaporation is 33455.16 g/h, which is equivalent to 33.5 kg/h.

To determine the rate at which water must be evaporated to crystallize 60% of the KOH in the feed, we can use the information given in the question.

1. First, let's calculate the moles of H2SO4 used in the titration:
- The molar concentration of H2SO4 is 0.85 mol/L.
- The volume of H2SO4 used is 22.4 mL, which is equivalent to 0.0224 L.
- Therefore, the moles of H2SO4 used is 0.85 mol/L x 0.0224 L = 0.01904 mol.

2. Next, let's determine the moles of KOH in the 5 g aliquot of the feed solution:
- The molar mass of KOH is 39.1 g/mol + 16.0 g/mol + 1.0 g/mol = 56.1 g/mol.
- The mass of KOH in the 5 g aliquot is 60% of 5 g, which is 0.6 x 5 g = 3 g.
- Therefore, the moles of KOH in the aliquot is 3 g / 56.1 g/mol = 0.053 mol.

3. Now, let's calculate the moles of KOH in the entire feed solution:
- The rate of feeding the solution is 875 kg/h, which is equivalent to 875,000 g/h.
- The concentration of KOH in the solution is 103 kg KOH/100 kg H2O, which is equivalent to 103 g KOH/100 g H2O.
- Therefore, the moles of KOH in the feed solution is (875,000 g / 100 g H2O) x (103 g KOH / 56.1 g/mol) = 1548.85 mol.

4. Now we can determine the moles of KOH that needs to be crystallized:
- The moles of KOH that need to be crystallized is 60% of the moles of KOH in the feed solution, which is 0.6 x 1548.85 mol = 929.31 mol.

5. Finally, let's calculate the rate of water evaporation:
- Since 2 moles of water are required for each mole of KOH in the crystal, the moles of water that need to be evaporated is 2 x 929.31 mol = 1858.62 mol.
- The rate of water evaporation is the moles of water evaporated per unit time. Since we are given the rate of feeding the solution in kg/h, we need to convert the moles of water to kg.

In summary, to crystallize 60% of the KOH in the feed, water needs to be evaporated at a rate of 33.5 kg/h.

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