For example, 1) pH of 12, and a pOH of 2 2) pH of 3, and a pOH of 11 Wouldn't the solution with the highest concentration of hydrogen ions be #1 and the highest concentration of hydroxide ions be #2. If not, please explain to me why. The higher the pH, the more hydroxide ions. PH measures the presence of H+ ions in solution. Since pH + POH = 14, the higher the pH, the lower the pOH and thus the higher concentration of OH in solution. This is because p means -log of. The answer is E.
Which of the following aqueous solutions has the lowest #[OH^-]# ?
A. a solution with a pH of 3.0
B. a #1*10^-4# M solution of #HNO_3#
C. a solution with a pOH of 12.0
D. pure water
E. a #1*10^-6#M solution of NaOH
2 Answers
Explanation:
Now in aqueous solution, under standard conditions, the following equilibrium operates...
#2H_2O(l) rightleftharpoons H_3O^+ + HO^-#
And #K_w=[H_3O^+][HO^-]=10^-14#...and if we take #log_10# of both sides..we get...
#14=pH+pOH#
And so we gots... #A. [H_3O^+]=0.001*mol*L^-1#
#B. [H_3O^+]=0.0001*mol*L^-1#
#C. pOH=12,#...so #pH=2#, and thus #[H_3O^+]=10^-2*mol*L^-1-=0.01*mol*L^-1#
#D.##pH=7# and thus #[H_3O^+]=10^-7*mol*L^-1#
#E.##[HO^-]=10^-6*mol*L^-1#, #pOH=+6#, #pH=8#, and so #[H_3O^+]=10^-8*mol*L^-1#
And so the HIGHEST concentration with respect to #H_3O^+# NECESSARILY demands the LOWEST CONCENTRATION with respect to #HO^-#. Note that we use #H_3O^+# and #H^+# interchangeably...
Capisce?
Explanation:
Let's examine the options by looking at their #'pH'#.
#A.# has a #'pH'=3#.
For #B.#, the #'pH'# of the nitric acid would be #-log[10^-4]=4#, as it is an acid, and so the equation corresponds to the #'pH'#.
For #C.#, the #'pH'# of the solution would be #14-'pOH'=14-12=2#.
#D.# is pure water, and therefore has a #'pH'=7#.
For #E.# the #'pOH'# of the sodium hydroxide would be #-log[10^-6]=6#, as it is a base, and so the logarithm equation responds to the #'pOH'#.
Since the #'pH'# is related to #'pOH'# by the equation,
#'pH + pOH'=14# at #25^@'C'#, we can find the #'pOH'# of the following choices.
#A.# has a #'pOH'=14-3=11#.
#B.# has a #'pOH'=14-4=10#.
#C.# has a #'pOH'=12#, as stated.
#D.# has a #'pOH'=14-7=7#.
#E.# has a #'pOH'=6#, as calculated.
Notice how,
Which Solution Below Has The Highest Concentration Of Hydroxide Ions Formed
#6<7<10<11<12#
or
#E.<D.<B.<A.<C.#
Since #'pOH'=-log[OH^-]#, or #[OH^-]=10^(-'pOH')#, and so the hydroxide ion concentration is inversely proportional to the #'pOH'#, i.e. more #[OH^-]# means less #'pOH'#, and vice-versa.
Which Solution Below Has The Highest Concentration Of Hydroxide Ions Chegg
Since #C.# has the highest #'pOH'#, then that means that it has the least #[OH^-]# concentration, and so it is the answer.
Which Solution Below Would Have The Highest Concentration Of Hydroxide Ions
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