6B11 part b

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annabelchen2a
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Joined: Wed Sep 30, 2020 10:01 pm

6B11 part b

Postby annabelchen2a » Mon Jan 18, 2021 12:49 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.
(b) What mass of Na2O was added to the first flask?

I got the molar concentration of hydroxide ions in the original solution to be 17.78M.
The answer key says the reaction is Na2O(s) + H2O(l) --> 2NaOH(aq) and the mass of Na2O is: 18mol/L * 0.2L * (1mol Na2O/2mol NaOH) * (61.98g Na2O/1mol Na2O) = 110g Na2O

I'm just confused as to why it's 18mol/L * 0.2L ... instead of 2(18)mol/L * 0.2L ... since I thought because the solution is 2NaOH and not NaOH, we would have to take the molar concentration of OH and multiply that by 2. Why is this wrong?

Samantha Pedersen 2K
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Re: 6B11 part b

Postby Samantha Pedersen 2K » Mon Jan 18, 2021 1:03 pm

I believe this is because the problem gives us the pH which already takes into account the fact that there are 2 moles of OH- in the solution. We use the pH to calculate the molar concentration of hydroxide ions in the diluted solution for i of Part A, then we use the resulting value to calculate the molar concentration of hydroxide ions in the original solution for ii of Part A. We use the value we obtained for ii of Part A in Part B. Therefore, we are using extensions of the pH in every part of the problem, and the pH already takes into account the fact that there are 2 moles of OH- in the solution. I hope this helped!

annabelchen2a
Posts: 101
Joined: Wed Sep 30, 2020 10:01 pm

Re: 6B11 part b

Postby annabelchen2a » Mon Jan 18, 2021 9:02 pm

Oh, that makes sense, thank you!


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