6B.11 part a ii

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Malia Shitabata 1F
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6B.11 part a ii

Postby Malia Shitabata 1F » Sun Jan 19, 2020 3:09 pm

A student added solid Na2O to a volumetric flask of volume 200.0 mL which was then filled with water resulting in 200.0 mL of NaOH solution. Then 5.00 mL of the solution was transferred to another volumetric flask and diluted to 500.0 mL. The pH of the diluted solution is 13.25. What is the molar concentration of hydroxide ions in the original solution?

Can somebody explain the math for this part? I understand part i, but I don't get why they multiplied by the 500.0 mL volume and divided by 5.00 mL.

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Re: 6B.11 part a ii

Postby VPatankar_2L » Sun Jan 19, 2020 3:33 pm

You can use M1V1 = M2V2 where M1 is the concentration of the original solution, M2 is 0.178 mol/L from part i, V2 is 500 mL, and V1 is 5 mL. You are using 5mL for V1 because that is the volume of the original solution that is used to dilute it to 500 mL.

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