You add 50.00 uL of 25.00 M NaOH (m = 39.9971 g/ mol) to 100.0 mL of a 0.252 mol/L NaCI (7=57.959 g/ mol) solution. The pH of the solution is: The pOH of the solution is: The pa of the solution is:
You add 50.00 uL of 25.00 M NaOH (m = 39.9971 g/ mol) to 100.0 mL of a 0.252 mol/L NaCI (7=57.959 g/ mol) solution. The pH of the solution is: The pOH of the solution is: The pa of the solution is:
Chapter14: Principles Of Neutralization Titrations
Section: Chapter Questions
Problem 14.30QAP
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You add 50.00 uL of 25.00 M NaOH (m = 39.9971 g/ mol) to 100.0 mL of a 0.252 mol/L
NaCI (7=57.959 g/ mol) solution.
The pH of the solution is:
The pOH of the solution is:
The pa of the solution is:
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