12.5 mM HCl and 7.5 mM KOH
after netralization ... 5.0 mM HCl in
75.0 mL of solution ... .067 M HCl
pH = -log[H⁺] = -log[.067]
50.00mL of 0.250M hydrochloric acid was added to 25.00mL of 0.300M potassium hydroxide.
Calculate the pH of the resulting solution.
1 answer