Answer:
a. pH = 2.04
b. pH = 3.85
c. pH = 8.06
d. pH = 11.56
Explanation:
The nitrous acid is a weak acid (Ka = 5.6x10⁻⁴) that reacts with NaOH as follows:
HNO₂ + NaOH → NaNO₂(aq) + H₂O(l)
a. At the beginning there is just a solution of 0.12M HNO₂. As Ka is:
Ka = [H⁺] [NO₂⁻] / [HNO₂]
Where [H⁺] and [NO₂⁻] ions comes from the same equilibrium ([H⁺] = [NO₂⁻] = X):
5.6x10⁻⁴ = X² / 0.15M
8.4x10⁻⁵ = X²
X = [H⁺] = 9.165x10⁻³M
As pH = -log [H⁺]
<h3>pH = 2.04</h3><h3 />
b. At this point we have HNO₂ and NaNO₂ (The weak acid and the conjugate base), a buffer. The pH of a buffer is obtained using H-H equation:
pH = pKa + log [NaNO₂] / [HNO₂]
<em>Where pH is the pH of the buffer,</em>
<em>pKa is -log Ka = 3.25</em>
<em>And [NaNO₂] [HNO₂] could be taken as the moles of each compound.</em>
<em />
The initial moles of HNO₂ are:
0.100L * (0.15mol / L) = 0.015moles
The moles of base added are:
0.0800L * (0.15mol / L) = 0.012moles
The moles of base added = Moles of NaNO₂ produced = 0.012moles.
And the moles of HNO₂ that remains are:
0.015moles - 0.012moles = 0.003moles
Replacing in H-H equation:
pH = 3.25 + log [0.012moles] / [0.003moles]
<h3>pH = 3.85</h3><h3 />
c. At equivalence point all HNO2 reacts producing NaNO₂. The volume added of NaOH must be 100mL. That means the concentration of the NaNO₂ is:
0.15M / 2 = 0.075M
The NaNO₂ is in equilibrium with water as follows:
NaNO₂(aq) + H₂O(l) ⇄ HNO₂(aq) + OH⁻(aq) + Na⁺
The equilibrium constant, kb, is:
Kb = Kw/Ka = 1x10⁻¹⁴ / 5.6x10⁻⁴ = 1.79x10⁻¹¹ = [OH⁻] [HNO₂] / [NaNO₂]
<em>Where [OH⁻] = [HNO₂] = x</em>
<em>[NaNO₂] = 0.075M</em>
<em />
1.79x10⁻¹¹ = [X] [X] / [0.075M]
1.34x10⁻¹² = X²
X = 1.16x10⁻⁶M = [OH⁻]
pOH = -log [OH-] = 5.94
pH = 14-pOH
<h3>pH = 8.06</h3><h3 />
d. At this point, 5mL of NaOH are added in excess, the moles are:
5mL = 5x10⁻³L * (0.15mol / L) =7.5x10⁻⁴moles NaOH
In 100mL + 105mL = 205mL = 0.205L. [NaOH] = 7.5x10⁻⁴moles NaOH / 0.205L =
3.66x10⁻³M = [OH⁻]
pOH = 2.44
pH = 14 - pOH
<h3>pH = 11.56</h3>
; the reaction is product favored.
; the reaction is reactant favored.
; the reaction is in equilibrium