Textbook problem 6B.3 part b

[Simrah_Ahmed1J]

How would you use the information given to arrange a -log([H⁺]) calculation in part b of this problem?

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 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?

[Joyce Pang 2H]

I'd use M1V1=M2V2 to find the actual concentration of the solution.

H+ and CL- are in equal parts so it's a 1;1 ratio, meaning [H+] should be equal to [HCl]. Using the molarity found, use -log[H+]. Hope that helps :)

[Megan Lu 3D]

Hi! Like Joyce said above, I would also use the dilution equation M1V1 = M2V2 to find the actual concentration, getting (0.025)(200.0) = [HCl](250.0), where [HCl] = 0.020 mol/L. Because there is a 1:1 ratio between HCl and H3O+, [HCl] = [H3O+] = 0.020 mol/L. Then, we can use pH = -log[H3O+]. Hope this was helpful!

[MaiVyDang2I]

Hey, I would use M1V1 = M2V22 to find the concentration of HCl first. So 0.025 M x 200.0 mL = M2 x 250.0 mL, so M2 would be 0.02 M. Because HCl is a strong acid, so it will deprotonate completely. [HCl] would then be equal to [H³O₊]. pH = -log[H³O₊] = -log(0.02) = 1.7.