Answer:
They are both listed under group 11 on the periodic table and both are highly conductive of electricity
Explanation:
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Answer:
Approximately
, assuming that this acid is monoprotic.
Explanation:
Assume that this acid is monoprotic. Let
denote this acid.
.
Initial concentration of
without any dissociation:
.
After
of that was dissociated, the concentration of both
and
(conjugate base of this acid) would become:
.
Concentration of
in the solution after dissociation:
.
Let
,
, and
denote the concentration (in
or
) of the corresponding species at equilibrium. Calculate the acid dissociation constant
for
, under the assumption that this acid is monoprotic:
.
The pH of the sodium hydroxide (NaOH) solution at the given concentration of 0.000519 M is determined as 10.72.
<h3>What is pH of solution?</h3>
The pH of a solution is defined as the logarithm of the reciprocal of the hydrogen ion concentration [H+] of the given solution.
Concentration of the basic solution, [OH⁻] = 0.000519
pOH = -log[OH⁻]
pOH = -log[0.000519]
pOH = 3.28
<h3>pH of the solution</h3>
pH + pOH = 14
pH = 14 - pOH
pH = 14 - 3.28
pH = 10.72
Thus, the pH of the sodium hydroxide (NaOH) solution at the given concentration of 0.000519 M is determined as 10.72.
Learn more about pH here: brainly.com/question/26424076