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Postby sharonvivianv » Sun Jan 20, 2019 1:37 pm

HW Question:

A careless laboratory technician wants to prepare 200.0 mL of a 0.025 m HCl(aq) solution, but uses a 250.0-mL volumetric fl ask 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 simply calculated the pH from the information and got 1.6
However, I don't know how to get part b based on the information given.

eden tefera 2B
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Re: HW Q

Postby eden tefera 2B » Sun Jan 20, 2019 3:25 pm

You should repeat the same process of finding pH, but instead use the actual volume used, which was 250 mL.

Cole Doolittle 2K
Posts: 30
Joined: Fri Sep 28, 2018 12:20 am

Re: HW Q

Postby Cole Doolittle 2K » Sun Jan 20, 2019 3:33 pm

You have to dilute the solution by multiplying .025M(200ml/250ml). Then, take the negative log of that new concentration and you should get the correct pH.

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