In an experiment, 0.438 L of 0.152 M NaOH is diluted with water to a final volume of 3.00 L.What is the pH of the dilute solution?

Question

Quizplus · Accepted Answer

The initial moles of NaOH are calculated by multiplying the initial concentration (0.152 M) by the initial volume (0.438 L), giving 0.066576 moles. When diluted to 3.00 L, the new concentration is 0.022192 M. NaOH is a strong base, so [OH-] = 0.022192 M. The pOH is -log(0.022192) = 1.654, and the pH is 14 - 1.654 = 12.346.

In an Experiment, 0