Part 1:
HI is a strong acid, so it completely dissociates in water to form H+ ions. Therefore, [H+] = 0.0015 M.
pH = -log[H+]
pH = -log(0.0015)
pH = 2.82
Therefore, the pH of 0.0015 M HI solution is 2.82.
Part 2:
NaOH is a strong base, so it completely dissociates in water to form OH- ions. Therefore, [OH-] = 0.00015 M.
pH = 14 - pOH
pOH = -log[OH-]
pOH = -log(0.00015)
pOH = 3.82
pH = 14 - 3.82
pH = 10.18
Therefore, the pH of 0.00015 M NaOH solution is 10.18.
Calculate the pH of each aqueous solution.
Note: Reference the Fundamental constants table for additional information.
Part 1 of 2
0.0015 M HI. Be sure your answer has the correct number of significant figures.
pH =
X
Part 2 of 2
0.00015 M NaOH. Be sure your answer has the correct number of significant figures.
pH =
X
1 answer
1 answer