## 6th Edition 12.27

Celeste 1I
Posts: 44
Joined: Fri Sep 28, 2018 12:18 am

### 6th Edition 12.27

A careless laboratory technician wants to prepare 200mL of a 0.025M HCL solution, but uses a 250.0mL volumetric flask instead. a) What is the pH of the desired solution?b) what is the actual pH? I guess I'm confused on how to start this.

Clarissa Cabil 1I
Posts: 66
Joined: Fri Sep 28, 2018 12:19 am

### Re: 6th Edition 12.27

You can first visualize the reaction by writing a balanced chemical equation:
HCl --> H+ + Cl- (HCl is a strong acid so it completely dissociates in water)

You can calculate the pH of the desired solution straight away because they give you the concentration, which is 0.025 M
pH = -log (0.025M)

You can then calculate the pH of the actual solution by taking the - log of the concentration
pH = - log (200 mL x 0.025 M / 250 mL)
You set up this equation because you need to cancel out the volumes.

I hope this helps!

Grace Kim 1J
Posts: 60
Joined: Fri Sep 28, 2018 12:18 am
Been upvoted: 1 time

### Re: 6th Edition 12.27

For part a, you plug in 0.025 directly into the equation [pH= negative log (H3O+) ]

For part b, you use the dilution equation M1V1 = M2V2 (you are solving for M2 in this case) in order to find the actual concentration of HCl, which will then give you the H30+ concentration to plug into the equation.

You should get 1.6 for part a and 1.7 for part b.

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