We will abbreviate malonic acid CH2(CO2H)2, a diprotic acid, as H2A (pK1 = 2.847 and pK2 = 5.696). Find the pH in (a) 0.200 M H2A; and (b) 0.200 M NaHA.

a. pH  → 1.77

b. pH → 4.27

Explanation:

Malonic acid is a dyprotic acid. It releases two protons:

H₂A  +  H₂O →  H₃O⁺   +  HA⁻     Ka1

HA⁻   +  H₂O →  H₃O⁺   +  A⁻²          Ka2

Let's find the first pH:

We expose the mass balance:

Ca = [HA]  +  [HA⁻] + [A⁻²] = 0.2 M

We can not consider the [A⁻²] so → Ca =  [HA]  +  [HA⁻] = 0.2M

As the acid is so concentrated, we can not consider the HA- so:

Ca = [HA] = 0.2 M

Charge balance →  [H⁺]  =  [HA⁻]  + [OH⁻]

H₂A  +  H₂O →  H₃O⁺   +  HA⁻     Ka1

Ka = H₃O⁺ . HA⁻ / H₂A

We need the HA⁻ value to put on the charge balance. We re order the Ka expression:

HA⁻ = Ka . H₂A / H₃O⁺           (notice that H₃O⁺ = H⁺)

We replace:  H⁺¨ = Ka . H₂A / H⁺

(H⁺)² = Ka . Ca

H⁺ = √(Ka . Ca)         We determine Ka from pKa → 10²'⁸⁴⁷ = 1.42×10⁻³

H⁺ = √(1.42×10⁻³ . 0.2)  =  0.016866

- log [H⁺] = pH  → - log 0.016866 = 1.77

b. NaHA is the salt from the weak acid, where the HA⁻ works as an amphoterous, this means that can be an acid or a base:

HA⁻  +  H₂O  ⇄  A⁻²  +  H₃O⁺     Ka₂

HA⁻  +  H₂O ⇄  H₂A  +  OH⁻      Kb2

There is a formula than can predict the pH, so now we need to compare the Ka₂ and Kb₂ data.

Ka₂ = 10⁻⁵'⁶⁹⁶ = 2.01×10⁻⁶

Kb₂ = 1×10⁻¹⁴ / 1.42×10⁻³  = 7.04×10⁻¹²

So Ka₂ > Kb₂. In conclussion the pH will be acid.

[H⁺] = √(Ka1 .  Ka2)  →  [H⁺] = √ 1.42×10⁻³ . 2.01×10⁻⁶ = 5.34×10⁻⁵

- log 5.34×10⁻⁵  = pH → 4.27

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.