the ph of a naoh solution is 12.05. what is the concentration of [oh-] of the naoh solution? express your answer in standard not
2 answers:
Answer:
0.01 M
Explanation:
NaOH (sodium hydroxide) is a strong base, so it is completely dissociated into ions in aqueous solution:
NaOH ⇄ Na⁺ + OH⁻
So, we have OH⁻ ions in solution.
From the problem, we have the pH value. The pH is calculated as:
pH = -log [H⁺]
As NaOH is a base, we can also calculate the pOH:
pOH = -log [OH⁻]
Thus. we can use the relation between pH and pOH to calculate the pOH and then use it to calculate [OH⁻]:
pH + pOH = 14 ⇒ pOH = 14 - pH = 14 - 12.05 = 1.95
pOH = -log [OH⁻] ⇒ [OH⁻] = =
= 0.0112 M ≅ 0.01 M
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Answer:
The correct answer is option B.
Explanation:
Moles of = 40 mol
Moles of NaOH = 48 mol
According to reaction, 3 moles of NaOH reacts with 2 moles
Then ,48 moles of NaOH will reacts with:
of
Then ,40 moles of will reacts with:
of NaOH
As we can see that 48 moles of sodium will completey react with 32 moles of nitrogen tribromide.
Moles left after reaction = 40 mol - 32 mol = 8 mol
Hence, the is an excessive reagent.