Question

Question #889ec

Answer 1

<, =, Neutral, Basic.

Explanation:

1. <
   Since the molarity of either acid is the same, the moles of each acid are equal. That means the difference between their pH is determined solely on which acid dissociates more. Since the strong acid dissociates more (releases more `H^+` ions), it will have a lower pH.

2. =
   Since both acids are monoprotic (they only release one hydrogen) they will both take the same amount of `OH^-` to be neutralized.

3. 1. Neutral
      The conjugate acid/base of strong acids and bases aren't strong enough to ionize water, so the pH will be exactly 7.

4. Basic
   The conjugate base of the weak acid is strong enough to react with water to form `OH^-`, causing the pH to be basic.

Answer 2

Strong Acid + Strong Base Titration at eqv pt => pH = 7
Weak Acid + Strong Base Titration at eqv pt => pH >7

Explanation:

Mixing of equal volumes of equimolar acid and base solutions will react completely. The pH at the equivalence points will be a function of salt ion hydrolysis.

`0.10 M NaOH + 0.10 M HCl => 0.10 M NaCl + 0.10M HOH`
`0.10M NaCl => 0.10M Na^+` + 0.10M `Cl^-`

`Na^+ + HOH => No Reaction` => NaOH is a strong Electrolyte and will not form molecular base in water. It prefers to remain 100% ionized.

`Cl^- + HOH => No Reaction` => HCl is a strong Electrolyte and will not form molecular acid in water. It prefers to remain 100% ionized.
Since neither of the salt ions reacts with water, the pH of a strong acid + strong base titration at the equivalence point is exclusively dependent upon the autoionization of water.

`HOH + HOH <=> H_3O^ + + OH^-`

`[H_3O^+] = [OH^-] = 1.0 x 10^-7M` at `25^oC`

`pH = -log[H_3O^+] = -log[10^(-7)] = 7.00`

`0.10M NaOH + 0.10M HOAc => 0.10M NaOAc + HOH`
`0.10M NaOAc => 0.10M Na^+ + 0.10M OAc^_`
`Na^+ + HOH => No Reaction`
`OAc^- + HOH => HOAc + OH^-` => Hydrolysis of acetate ion => Acetic Acid + excess of Hydroxide ions. emphasized text This is what gives the Strong Base + Weak Acid an alkaline pH at the equivalence point.

We set up an ICE table as usual.

`color(white)(mmmmmmmmml)OAc^- + H_2O ⇌ HOAc + OH^-`

`"I/mol·L"^"-1":color(white)(mmmm)0.10color(white)(mmmmmmmll)0color(white)(mmmm)0`
`"C/mol·L"^"-1":color(white)(mmmml)"-"x"color(white)(mmmmmml)+xcolor(white)(mml)+x`
`"E/mol·L"^"-1":color(white)(mmml)"0.10-"x"color(white)(mmmmmmm)xcolor(white)(mmmm)x`

`(K_b) (K_a) = K_w` => `K_b = K_w/K_a = (1.0 x 10^(-14))/(1.8x10^(-5)) = 5.5x10^(-10)`

`K_b = (([HOAc][OH^-])/ "[OAc^-]") = ("(X)(X)"/(0.10)) = (X^2/0.10) = 5.5X10^(-10)`

=> `X = sqrt((5.5X10^(-10))(0.10))M = 7.45x10^(-6)M = [OH^-]`

=> `pOH = -log[OH^-] = -log(7.45x10^(-6)) = 5.13`

=> `pH + pOH = 14 => pH = 14 - pOH = 14 - 5.13 = 8.87`