What happens to the entropy when a solution is made?
A. The entropy increases.
B. The entropy decreases.
C. The entropy goes to zero.
D. The entropy is unaffected.

Answer:

b < c < a < d

Explanation:

The weak acid with the lowest pKa will be the most acidic. In the other way, the conjugate base which the acid is weak will be strong.

The weak base with the lowest pKb will be the most basic. And the conjugate base of the weak base will be a strong acid.

Ka Acetic acid = 1.8x10-5

Ka HCN = 1.9x10-10

Kb pyridine = 1.7x10-9

Kb NH3 = 1.8x10-5

NH4Br is the conjugate base of a weak base. That means is a strong acid.

NH4Br has the lowest pH

NaBr is the conjugate base of a strong acid, HBr. That means NaBr is neutral

The most basic between the conjugate base of the acetic acid, NaCH3CO2 and KCN is KCN because the acetic acid is the stronger acid regard to HCN.

The rank is:

NH4Br < NaBr < NaCH3CO2 < KCN

Answer:

Volume is reduced to half

Explanation:

Acid base titration are commonly used reactions in a lab, and are ofter used to get pH or different kind of solutions.

The neutralization of an acid base reaction is reached, when the solution (having added an indicator previously) changes its original color. chemically speaking, this occurs when the number of moles of the acid and the base are balanced and equal. In other words the following:

n₁ = n₂   (1)

This expression can also be expressed in function of concentration and volume:

M₁V₁ = M₂V₂   (2)

From here, solving for V₁:

V₁ = M₂V₂  / M₁

Now, this expression is true only when we have the same kind of substance that can lose or gain the same number of hydrogens.

Lets suppose that we have as base NaOH (Monoprotic base) and HCl (monoprotic acid), the titration reaction would be:

NaOH + HCl --------> NaCl + H₂O

As both of the species are monoprotic, the number of moles are the same  when they reach the equilibrium, so, expression  (2) can be used, and calculate volume or concentrations.

However, in this case, a partner made a mistake and use a diprotic acid, in this case, H₂SO₄, In this case, things chance because H₂SO₄ is diprotic, meaning that we need to dissociate two hydrogens in equilibrium, therefore, expression (2) would be something like this.

Acid: 1; Base: 2

H₂SO₄ + 2NaOH ------> Na₂SO₄ + H₂O

nH₂SO₄ = n₁ = 1

nNaOH = n₂ = 2

n₁/n₂ = 1/2

2n₁ = n₂   (3)

Writting this, in function of concentration and volume, it would be:

2M₁V₁ = M₂V₂   (4)

From here, if we solve for the volume of the acid (V₁):

V₁ = M₂V₂ / 2M₁

Therefore, according to this expression, we can see that the volume required of the acid would be half the volume required of the monoprotic acid. For example, if we need 50 mL of Chloridic acid to reach the equivalence point with NaOH, then, with H₂SO₄ it will only need 25 mL. This, of course, assuming that concentrations are the same, and volume of the base used, the same.

Hope this helps

Answer:

See explanation

Explanation:

Hello there!

In this case, since the the concentrations are not given, and not even the Ksp, we can solve this problem by setting up the chemical equation, the equilibrium constant expression and the ICE table only:

[tex]AgCl(s)\rightleftharpoons Ag^+(aq)+Cl^-(aq)[/tex]

Next, the equilibrium expression according to the produced aqueous species as the solid silver chloride is not involved in there:

[tex]Ksp=[Ag^+][Cl^-][/tex]

And therefore, the ICE table, in which x stands for the molar solubility of the silver chloride:

       [tex]\ \ \ \ \ \ \ \ \ \ \ \ \ \ AgCl(s)\rightleftharpoons Ag^+(aq)+Cl^-(aq)[/tex]

I          -                   0             0

C        -                   +x           +x

E        -                    x             x

Which leads to the following modified equilibrium expression:

[tex]Ksp=x^2[/tex]

Unfortunately, values were not given, and they cannot be arbitrarily assigned or assumed.

Regards!