Find the pH of a solution prepared by taking a 50.0 mL aliquot of a solution prepared by dissolving 12.25g of NaOH and 250.0 mL of water and diluting that aliquot to 1.00L?

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Question:

Find the pH of a solution prepared by taking a 50.0 mL aliquot of a solution prepared by dissolving 12.25g of NaOH and 250.0 mL of water and diluting that aliquot to 1.00L?

Answer:

Answer:

12.8

Explanation:

Considering:

?f=Molarity%3D%5Cfrac%7BMass%7D%7BMolar%20mass%5Ctimes%20Volume%5C%20of%5C%20the%5C%20solution%7D

For NaCl:-

Mass = 12.25 g

Molar mass = 58.44 g/mol

Volume of solution = 250.0 mL = 0.25 L

So,

?f=Molarity %7Bstock%5C%20solution%7D%3D%5Cfrac%7B12.25%7D%7B39.997%5Ctimes%200.25%7D%5C%20M%3D1

Considering

?f=Molarity %7Bworking%5C%20solution%7D%5Ctimes%20Volume %7Bworking%5C%20solution%7D%3DMolarity %7Bstock%5C%20solution%7D%5Ctimes%20Volume %7Bstock%5C%20solution%7D

Given  that:

?f=Molarity %7Bworking%5C%20solution%7D%3D%3F

?f=Volume %7Bworking%5C%20solution%7D%3D1

?f=Volume %7Bstock%5C%20solution%7D%3D0

?f=Molarity %7Bstock%5C%20solution%7D%3D1

So,  

?f=Molarity %7Bworking%5C%20solution%7D%5Ctimes%201.00%3D1.225%5Ctimes%200

Concentration of NaOH = Concentration of [OH⁻] = 0.06125 M

pOH = – log[OH⁻] = -log(0.06125) = 1.21

pH = 14 – pOH = 14 – 1.21 = 12.8

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