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### 6B.3 PART B HELP

Posted: Thu Jan 16, 2020 10:44 pm
Question : 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?

I solved the second part by converting the volume of the flask actually used into L (0.25 L). Then I divided the moles of the HCl of the solution by the volume of the flask [(0.025 m) / (0.25 L)] and got 0.1 m*L^-1, which I divided again by 0.25 L to get 0.4 m. I then solved for the pH using -log[H3O+] and got 0.4 as my pH, which is wrong according to the solutions manual.

I'm not sure where I went wrong, if anyone sees where I messed up, I'd appreciate the heads up! :)

### Re: 6B.3 PART B HELP

Posted: Thu Jan 16, 2020 11:01 pm
You first need to multiple .025M by the volume 200.0mL (or .200L), not divide. You want to multiply so you get the number of moles, which you can then divide by the new volume to find mol/L.

### Re: 6B.3 PART B HELP

Posted: Fri Jan 17, 2020 1:12 am
For part B, the solution has been diluted due to the mistake made by the technician, so in order to solve for the pH, you would need to know the new concentration of the solution. To find that, you need to use the dilution equation: M1V1=M2V2.

### Re: 6B.3 PART B HELP

Posted: Fri Jan 17, 2020 12:57 pm
Chem_Mod wrote:You first need to multiple .025M by the volume 200.0mL (or .200L), not divide. You want to multiply so you get the number of moles, which you can then divide by the new volume to find mol/L.

Ahhh, I see, I'll try that, thank you!

### Re: 6B.3 PART B HELP

Posted: Fri Jan 17, 2020 12:58 pm
Alice Ma 2K wrote:For part B, the solution has been diluted due to the mistake made by the technician, so in order to solve for the pH, you would need to know the new concentration of the solution. To find that, you need to use the dilution equation: M1V1=M2V2.

Interesting, it didn't cross my mind to use that formula. I'll try this, thank you!