X.
In NaOH solution [OH®] is 2.87 *
104. Calculate the pH of solution.
Answers
Explanation:
For starters, you know that an aqueous solution kept at room temperature has
pH
+
pOH
=
14
−−−−−−−−−−−−−−−−
so use this equation to find the pOH of the solution.
pOH
=
14
−
9.4
=
4.6
Now, the pOH of the solution is given by the concentration of hydroxide anions,
OH
−
pOH
=
−
log
(
[
OH
−
]
)
−−−−−−−−−−−−−−−−−−−−
To find the concentration of hydroxide anions starting from this equation, rewrite it as
log
(
[
OH
−
]
)
=
−
pOH
10
log
(
[
OH
−
]
)
=
10
−
pOH
this will get you
[
OH
−
]
=
10
−
pOH
→
remember this equation!
In your case, you have
[
OH
−
]
=
10
−
4.6
[
OH
−
]
=
2.5
⋅
10
−
5
M
Answer:10.46 is the ans.
Explanation:The hydroxide ion concentration of NaOH is 2.9x10-4 M. What is the pH of the solution?
Since NaOH is a strong base it will completely ionize. Therefore, if the molarity of the solution is 2.9x10^4 , the concentration of OH ions will be 2.9x10^-4.
The pOH of the solution is determined using the formula pOH = log 1/2.9x10^-4 which is 3.54
Now recall that the pH of a basic solution is Determined using the equation pH= 14-pOH
The pH of this NaOH solution is 24–3.54 = 10.46.
Something to remember: in chemistry, the letter “p” is a mathematical operator. Just as + says “add”, p says “take the log of the reciprocal of”
You want pH. Calculate [H+]
NaOH dissociates NaOH → Na + + OH-
[OH-] = [Na+] = [NaOH] = 2.9*10^-4M
Equation:
[H+] [OH-] = 1*10^-14
[H+] = 1*10^-14 / [OH-]
[H+] = 1*10^-14 / (2.9*10^-4)
[H+] = 3.45*10^-11M
pH = -log [H+]
pH = -log ( 3.45*10^-11)
pH = 10.46