### 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! :)

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! :)