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 free energy change for the reaction at 37.0°C is -8.741 kJ.
Explanation:
The free energy of the reaction is given by :
where,
= standard Gibbs free energy
R = Gas constant =
T = temperature in Kelvins
K = equilibrium constant
We have :
1 kJ = 1000 J
T = 37.0 C = 37 +273.15 K = 310.15 K
Ratio of concentrations of the products to the concentrations of the reactants =K = 21.9
The free energy change for the reaction at 37.0°C is -8.741 kJ.