Question

What will be the PH of 0.002 MHcl
How do you arrive at o. 301

Answer 1

Answer:

pH=−log10[H3O+

Step-by-step explanation:

[H3 O+] = 10−2⋅L×0.1⋅mol⋅L−1500⋅mL×10−3L⋅ mL−1. = 2.0×10−3mol⋅L−1.

Missing: 0.301 ‎| Must include: 0.301