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.
The answer is E - have both hydrophilic and hydrophobic groups.
Based on the charge on the aluminium ion, 0.9 g of aluminium are deposited by 0.1 F of electricity.
<h3>What is electrolysis?</h3>
Electrolysis is the decomposition of a substance known as an electrolyte when electric current is passed through it.
The mass and hence moles an electrolyte deposited when current is passed through it depends on the charge on the ion.
Aluminium ion has a charge of +3 and requires 3F of electricity to deposit 1 mole or 27 g of aluminium
0.1 F will discharge = 0.1/3 × 27 g of aluminium
mass of aluminium deposited = 0.9 g of aluminium.
Therefore, 0.9 g of aluminium are deposited by 0.1 F of electricity.
Learn more about electrolysis at: brainly.com/question/26050361
Option a) H-H is the correct answer
