6 B.3

[Hamza M 3I]

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. What would the pH of the desired solution have been?

[Ambar Zamuria 1E]

For both parts (a) and (b), use the equation: pH=-log[H3O+], since [H3O+]=[HCl]. Just for part (b), you need to find the actual [HCl] before plugging it in [H3O+].

[Michelle Fong 3F]

To start this problem, you can start by finding the number of moles of HCl by converting 200.0 mL to L. You can multiply the molarity by the volume to get the number of moles of HCl. Because HCl is a strong acid, we know that it fully dissociates in water so we know that the concentration of hydronium ions is equal to the original concentration of HCl. The pH of the intended solution would therefore be equal to -log(0.025). To find the actual concentration, divide the number of moles of HCl by the volume of the 250.0 mL flask used after converting it to liters. You can use this value to calculate the pH of the actual solution.