Answer:
pH =3.8
Explanation:
Lets call the monoprotic weak acid HA, the dissociation equilibria in water will be:
HA + H₂O ⇄ H₃O⁺ + A⁻ with Ka = [ H₃O⁺] x [A⁻]/ [HA]
The pH is the negative log of the H₃O⁺ concentration, we know the equilibrium constant, Ka and the original acid concentration. So we will need to find the [H₃O⁺] to solve this question.
In order to do that lets set up the ICE table helper which accounts for the species at equilibrium:
HA H₃O⁺ A⁻
Initial, M 0.40 0 0
Change , M -x +x +x
Equilibrium, M 0.40 - x x x
Lets express these concentrations in terms of the equilibrium constant:
Ka = x² / (0.40 - x )
Now the equilibrium constant is so small ( very little dissociation of HA ) that is safe to approximate 0.40 - x to 0.40,
7.3 x 10⁻⁶ = x² / 0.40 ⇒ x = √( 7.3 x 10⁻⁶ x 0.40 ) = 1.71 x 10⁻³
[H₃O⁺] = 1.71 x 10⁻³
Indeed 1.71 x 10⁻³ is small compared to 0.40 (0.4 %). To be a good approximation our value should be less or equal to 5 %.
pH = - log ( 1.71 x 10⁻³ ) = 3.8
Note: when the aprroximation is greater than 5 % we will need to solve the resulting quadratic equation.
Answer:<span> a) the process for the first ionization energy
</span>
<span>Every time you take a electron you're requiring more and more energy. Expelling the first one
will require less energy than expelling the second and the second will
require less than the third, and so on.
When you take the first one, the atom becomes positive and with that the negative forces of the electron will be more attracted to the positive
charge. The more electrons that are lost, the
more positive this ion will become, causing it to be more difficult to separate the
electrons from the atom.
</span>
Answer:
0.978 M
Explanation:
Given data
- Mass of luminol (solute): 13.0 g
- Volume of the solution = volume of water: 75.0 mL = 0.0750 L
We can find the molarity of the stock solution of luminol using the following expression.
M = mass of solute / molar mass of solute × liters of solution
M = 13.0 g / 177.16 g/mol × 0.0750 L
M = 0.978 M
Answer:
Explanation:
The given pH = 8.55
Unknown:
[H₃O⁺] = ?
[OH⁻] = ?
In order to find these unknowns we must first establish some relationship.
pH = -log[H₃O⁺]
8.55 = -log[H₃O⁺]
[H₃O⁺] = inverse log₁₀(-8.55) = 2.82 x 10⁻⁹moldm⁻³
To find the [OH⁻],
pH + pOH = 14
pOH = 14 - pH = 14 - 8.55
pOH = 5.45
pOH = -log[OH⁻]
[OH⁻] = inverse log₁₀ (-5.45) = 3.55 x 10⁻⁶moldm⁻³
The solution is basic because it has more concentration of OH⁻ ions compared to H⁺ ions.
