Chemistry

How many millimeters of a 15.0%, by mass solution of KOH(aq)(d = 1.14 g/ml ) are required to produce 26.0 L of a solution with pH = 11.40?

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1. A 15% solution is 0.15 g KOH/g solution. And since the density is 1.14 g/mL, that is 0.15g/1.14 mL or about 0.13 g/1 mL. How much of that do you need?
pH = 11.4. Convert to pOH and from there to (OH^-). I get close to 2.5 M. That is 2.5 mols KOH/liter of solution. How much do you need? You want to make up 26.0 L therefore, that is 2.5 moles KOH x 26.0 L = approximately 0.06 mols and that x molar mass KOH = about 3 grams KOH to make up to 26.0 L.
So set up a proportion
0.13 g KOH/1 mL KOH = 3.0 g KOH/? mL)
I get something like 26 mL or so. You need to go through and use more exact figures since I estimated here and there. Check my work.

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2. I thought for the density, the calculation would be as follows: 0.15gKOH/g---> 1g of solution, density=1.14 g/ml.......0.15g/(1g/1.14g/ml) to get 0.171 g/ml. and then I thought molarity for [OH^-] was 2.51 * 10^-3. The proper set up I got like 3.66 g/? mL. I might be wrong!! plzz can u check what I did and have a final say on the answer

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