12.27

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Abigail Urbina 1K
Posts: 102
Joined: Thu Jul 27, 2017 3:01 am

12.27

Postby Abigail Urbina 1K » Mon Dec 04, 2017 10:04 pm

I don't understand how to solve part b of #27 on the chapter 12 homework. Can someone explain how they went about to setup the calculation for part b? I didn't quite get how they did it in the solution manual

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 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?

Angel Ni 2K
Posts: 50
Joined: Sat Jul 22, 2017 3:01 am

Re: 12.27

Postby Angel Ni 2K » Mon Dec 04, 2017 10:25 pm

From the problem, we know that the concentration of HCl in the intended solution is 0.025M. Therefore, [H3O+] in the intended solution is also 0.025M. The number of moles of HCl remains the same between intended and actual solutions. From earlier in the course, we learned that M1V1=M2V2. Thus, using (0.025M)(0.2000L)=M2(0.2500L), we find that the actual [H3O+] is 0.020M. Take the negative log of 0.020 and you will find that the pH is 1.7


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