6th edition, 12.33

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Tyra Nguyen 4H
Posts: 74
Joined: Fri Sep 28, 2018 12:25 am

6th edition, 12.33

Postby Tyra Nguyen 4H » Wed Jan 23, 2019 10:05 am

The question is as follows:
A student added solid Na2O to a 200.0-mL volumetric flask, which was then filled with water, resulting in 200.0 mL of NaOH solution. 5.00 mL of the solution was then transferred to another volumetric flask and diluted to 500.0 mL. The pH
of the diluted solution is 13.25. What is the concentration
of hydroxide ion in (a) the diluted solution? (b) the original solution? (c) What mass of Na2O was added to the first flask?

How do you solve part c? I utilized the moles calculated from part a/b and attempted to calculate the mass on that because of the stoichiometric coefficients, but I'm got an answer that was 20x too small.

ChathuriGunasekera1D
Posts: 78
Joined: Fri Sep 28, 2018 12:17 am

Re: 6th edition, 12.33

Postby ChathuriGunasekera1D » Wed Jan 23, 2019 10:31 am

Hi! So using the 17.78 M OH from part b, you can use that and the balanced equation to work backwards. Because NaOH is a strong base, it will completely dissociate, meaning that [OH] = [Na] = [NaOH]. If we have 17.78 M [OH], the we have 17.78 M [NaOH].

The equation is Na2O + H2O ----> 2NaOH
If we have 17.78 M NaOH in 200 mL, we have 3.556 mol of NaOH. Because we have 2 moles of NaOH for every mole of Na2O, that means we have half as many moles of Na2O, which is 3.556/2 moles of Na2O (1.778 moles). If we convert this to grams, it's 1.778 moles x 62 g/moles which gives about 110 g.

Tyra Nguyen 4H
Posts: 74
Joined: Fri Sep 28, 2018 12:25 am

Re: 6th edition, 12.33

Postby Tyra Nguyen 4H » Wed Jan 23, 2019 10:35 am

ChathuriGunasekera1D wrote:Hi! So using the 17.78 M OH from part b, you can use that and the balanced equation to work backwards. Because NaOH is a strong base, it will completely dissociate, meaning that [OH] = [Na] = [NaOH]. If we have 17.78 M [OH], the we have 17.78 M [NaOH].

The equation is Na2O + H2O ----> 2NaOH
If we have 17.78 M NaOH in 200 mL, we have 3.556 mol of NaOH. Because we have 2 moles of NaOH for every mole of Na2O, that means we have half as many moles of Na2O, which is 3.556/2 moles of Na2O (1.778 moles). If we convert this to grams, it's 1.778 moles x 62 g/moles which gives about 110 g.


So the OH concentration is from the OH in NaOH? I initially had the same equation as you, but got confused and rewrote the equation as Na2O+H2O --> NaOH + OH + Na.

ChathuriGunasekera1D
Posts: 78
Joined: Fri Sep 28, 2018 12:17 am

Re: 6th edition, 12.33

Postby ChathuriGunasekera1D » Fri Jan 25, 2019 7:13 am

Yes! The OH comes from the NaOH. If you wanted to write another equation, you could have Na2O+H2O --> OH + Na to show the inevitable dissociation of NaOH into Na and OH, but not Na2O+H2O --> NaOH + OH + Na because it kind of implies that NaOH does not completely dissociate.


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