6B.3

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Chris Tai 1B
Posts: 102
Joined: Sat Aug 24, 2019 12:16 am

6B.3

Postby Chris Tai 1B » Tue Dec 03, 2019 5:52 pm

A careless laboratory technician wants to prepare 200.0 mL of a 0.025 m HCl(aq) solution but uses a volumetric flask of volume 250.0 mL by mistake. (a) What would the pH of the desired solution have been? (b) What will be the actual pH of the solution as prepared?

For part a, I started by multiplying .2 L by 0.025 M to get 0.005 mols HCl, so I assumed there would be 0.005 mols hydronium. I get 2.3 as my final pH (the right answer should be 1.6) - how should I have approached the problem differently to get the correct answer?

Trinity Vu 1D
Posts: 53
Joined: Fri Aug 30, 2019 12:15 am

Re: 6B.3

Postby Trinity Vu 1D » Tue Dec 03, 2019 5:57 pm

In order to calculate pH you use the concentration of hydronium. Therefore you need to divide mol H3O by L of solution which will give you .025 M H3O. Same for (b) You divide the .005 mol H30 by .250L in order to get the new concentration ([H3O] which gives you .02. Then you can take the -log of this which gives you 1.7.

Letty Liu 2C
Posts: 105
Joined: Fri Aug 09, 2019 12:16 am

Re: 6B.3

Postby Letty Liu 2C » Wed Dec 04, 2019 9:38 pm

Trinity Vu 1D wrote:In order to calculate pH you use the concentration of hydronium. Therefore you need to divide mol H3O by L of solution which will give you .025 M H3O. Same for (b) You divide the .005 mol H30 by .250L in order to get the new concentration ([H3O] which gives you .02. Then you can take the -log of this which gives you 1.7.


How do you know how many mol H30 there are to divide it by the L of solution? Is it just .250 because the moles of product is equal to the moles of the reactants?


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