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Answer:
The [OH⁻] of a solution that has a pOH of 2.7 will be 2*10⁻³
Explanation:
The pOH (or potential OH) is a measure of the basicity or alkalinity of a solution.
pOH indicates the concentration of hydroxyl ions [OH-] present in a solution. In this way, pOH is defined as the negative logarithm of the activity of hydroxide ions, that is, the concentration of OH- ions:
pOH= -log [OH⁻]
In this case, pOH has a value of 2.7. Replacing:
2.7= -log [OH⁻]
and solving:
[OH⁻]=10⁻² ⁷
you get:
[OH⁻]≅ 2*10⁻³
<u><em>The [OH⁻] of a solution that has a pOH of 2.7 will be 2*10⁻³</em></u>
Answer: PH= 1.4
Explanation:
n(mole)= concentration(c)× volume(v)
n(HClO4) = 0.08 ×0.03
n(HClO4)= 0.0024 mole
n(Ba(OH)2) = 0.28 × 0.05
n(Ba(OH)2) = 0.014
2HClO4 + Ba(OH)2.....>Ba(ClO4)2 + 2H2O.
Hence, the limiting reagent is perchloric acid. According to the reaction:
0.0024 HClO4 req 1/2× 0.0024 of Ba(OH)2.
n(Ba(OH) reacted= 0.0012
Excess n(Ba(OH)2) = 0.014-0.0012
Excess n(Ba(OH)2) = 0.0128
Conc(Ba(OH)2) unreacted = 0.0128/.32 (vol. of mix is 32ml)
= 0.04M = [H+]
PH = -log[0.04] = 1.397
PH= 1.4
Answer C
I could be wrong
Explanation
