Part 1:
Given: 0.000024 M HI
Since HI is a strong acid, it completely dissociates in water to form H+ ions and I- ions.
[H+] = 0.000024 M
pH = -log[H+]
pH = -log(0.000024)
pH = 4.62
Therefore, the pH of 0.000024 M HI solution is 4.62.
Part 2:
Given: 0.000032 M KOH
Since KOH is a strong base, it completely dissociates in water to form K+ ions and OH- ions.
[OH-] = 0.000032 M
pOH = -log[OH-]
pOH = -log(0.000032)
pOH = 4.50
Since pH + pOH = 14,
pH = 14 - pOH
pH = 14 - 4.50
pH = 9.50
Therefore, the pH of 0.000032 M KOH solution is 9.50.
help me with this Calculate the pH of each aqueous solution.
Note: Reference the Fundamental constants table for additional information.
Part 1 of 2
0.000024 M HI. Be sure your answer has the correct number of significant figures.
pH =
Part 2 of 2
0.000032 M KOH. Be sure your answer has the correct number of significant figures.
pH =
1 answer
1 answer