6B. 3

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Janet Nguy 2C
Posts: 100
Joined: Sat Aug 24, 2019 12:17 am

6B. 3

Postby Janet Nguy 2C » Sun Jan 19, 2020 3:47 pm

6B.3 A careless laboratory technician wants to prepare 200.0 mL
of a 0.025 m HCl(aq) solution but uses a volumetric flask of vol-
ume 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?


I did this problem but I don't have the solutions manual so I can't confirm that I did it right. Can someone take a look at my thinking and verify it for me? Here's my thought process:

a)
Balanced equation: H20(l) + HCl(aq) > H30+(aq) + Cl-(aq)
HCl is a strong acid, so it is completely dissociated and this implies that having 0.025M HCl yields 0.025M H30+.
pH = -log(H30+) = -log(0.025) = 1.60

b)
I assume they want me to use M1V1 = M2V2 because changing the volume of the flask also changes the molarity of HCl.
Therefore, (0.025 M HCl)(0.2 L) = M2(0.25 L)
M2 = 0.02 M HCl in the 250 mL flask
so pH = -log(0.02) = 1.70

Jeremy_Guiman2E
Posts: 68
Joined: Fri Sep 28, 2018 12:29 am

Re: 6B. 3

Postby Jeremy_Guiman2E » Sun Jan 19, 2020 4:49 pm

This is how I did it, and your process looks good!

Reina Robles 2B
Posts: 61
Joined: Fri Aug 09, 2019 12:16 am

Re: 6B. 3

Postby Reina Robles 2B » Tue Jan 21, 2020 12:32 am

I have the solutions manual and can also confirm that it's correct!


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