HW 12.33(c)

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Ashley Van Belle 2B
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HW 12.33(c)

Postby Ashley Van Belle 2B » Wed Nov 30, 2016 10:50 pm

A student added solid Na2O to a 200.0-mL 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 flask and diluted to 500.0mL. The pH of the diluted solution is 13.25. What is the concentration of OH- in (a) the diluted solution? (b) the original solution? (c) What mass of Na2O was added to the first flask?

When I approached this problem, I took the 18 mol/L concentration of OH- calculated from (b) and multiplied it by the .200 L of the original and by the molar mass: (18 mol*L-1)(0.200 L)(61.98g Na2O/1mol Na2O) to get about 223 grams.

However, when I looked at the solutions, it included (1mol Na2O/2mol NaOH) with the calculation: (18 mol*L-1)(0.200 L)(61.98g Na2O/1mol Na2O)(1mol Na2O/2mol NaOH). But why is this the case? NaOH would not be canceled out with the units, so I am confused why it would be necessary to include that information.

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Re: HW 12.33(c)

Postby Chem_Mod » Fri Dec 02, 2016 11:06 am

18 M is the concentration of OH-.

Na2O + H2O -> 2NaOH


Therefore to solve for moles of Na2O we need to take stoichiometry into consideration, which is 1:2.


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