Question

What is the pH of the solution obtained by mixing 30.00 mL of 0.250 M HCl...

What is the pH of the solution obtained by mixing 30.00 mL of 0.250 M HCl and 30.00 mL of 0.125 M NaOH?

Homework Answers

Answer #1

HCl + NaOH = H2O + NaCl

- Moles of H+ from the HCl = 0.250 moles/liter x 0.030 L = 0.0075 moles H+
- Moles of OH- from the NaOH = 0.125 moles/liter x 0.030 L = 0.00375 moles OH-

Now, you can subtract whichever number of moles is smaller from the larger number of moles. Here it's:

0.0075 moles H+ - 0.00375 moles OH- = 0.00375 moles H+ left over.

Don't forget to turn this back into molarity Right now you only have moles of H+...

You divide 0.00375 moles by the TOTAL volume of your solution in liters. Here that number is:

0.030 L + 0.030 L = 0.060 L total volume.

So,

0.00375 moles/ 0.060 Liters = 0.0625 M H+

pH = -log[H+]
-log(0.0625) = 1.204

pH = 1.20

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