Lulu Labwrecker carefully pipets 25.0 mL of 0.525 M NaOH into a test tube. She places the test tube into a small beaker to keep it from spilling and then pipets 75.0 mL of 0.355 M HCl into another test tube. When Lulu reaches to put this test tube of acid into the beaker along with test tube of base she accidentally knocks the test tubes together hard enough to break them and their respective contents combine in the bottom of the beaker. Is the solution formed from the contents of the two test tubes acidic or basic? What is the pH of the resulting solution?

Question

Chemistry

Elise Boyd

3 January, 20:36

Lulu Labwrecker carefully pipets 25.0 mL of 0.525 M NaOH into a test tube. She places the test tube into a small beaker to keep it from spilling and then pipets 75.0 mL of 0.355 M HCl into another test tube. When Lulu reaches to put this test tube of acid into the beaker along with test tube of base she accidentally knocks the test tubes together hard enough to break them and their respective contents combine in the bottom of the beaker. Is the solution formed from the contents of the two test tubes acidic or basic? What is the pH of the resulting solution?

+4

Answers (1)

Know the Answer?

Jessica · Answer 1

The solution formed is acidic

pH = 0.87

Explanation:

The acid-base reaction of NaOH with HCl is:

NaOH + HCl → NaCl + H₂O

Where 1 mole of NaOH reacts with 1 mole of HCl

In the problem, moles of NaOH and HCl are:

NaOH: 0.0250L * (0.525mol / L) = 0.013125 moles NaOH

HCl: 0.0750 * (0.355mol / L) = 0.026625 moles HCl

As moles of HCl > moles NaOH, HCl is in excess and the solution formed is acidic

Moles in excess of HCl are:

0.026625 moles - 0.013125 moles = 0.0135 moles HCl

As the volume of the solutions is 25.0mL + 75.0mL = 100.0mL = 0.100L, molarity of HCl after reaction is:

0.0135 moles HCl / 0.100L = 0.135M HCl = 0.135M H⁺

As pH is defined as - log [H⁺], pH of the solution is:

pH = - log 0.135M H⁺ = 0.87

pH = 0.87