Asked by Tommy
50.00mL of 0.250M hydrochloric acid was added to 25.00mL of 0.300M potassium hydroxide.
Calculate the pH of the resulting solution.
Calculate the pH of the resulting solution.
Answers
Answered by
Scott
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]
after netralization ... 5.0 mM HCl in
75.0 mL of solution ... .067 M HCl
pH = -log[H⁺] = -log[.067]
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