Answer:
pH = 9.37
Step-by-step explanation:
The reaction of HZ with NaOH occurs as follows:
HZ + NaOH → H₂O + Na⁺ + Z⁻
When the reaction is in the equivalence point, the moles added of HZ = Moles of NaOH.
At equivalence point, you will have in solution just NaZ that is in equilibrium with water as follows:
Z⁻(aq) + H₂O(l) ⇄ OH⁻(aq) + HZ(aq)
Where K of equilibrium is Kw / Ka = 1x10⁻¹⁴ / 2.4x10⁻⁶ = 4.17x10⁻⁹
And K is defined as:
K = 4.17x10⁻⁹ = [OH⁻] [HZ] / [Z⁻]
To solve this question we must find [Z⁻] and as [OH⁻] and [HZ] comes from the same equilibrium, both are equal and we can find [OH⁻] in order to solve pH:
Moles NaOH added = Moles HZ:
0.010L * (0.39mol / L) = 3.9x10⁻³ moles HZ = Moles NaOH added
The volume of 0.200M NaOH must be:
3.9x10⁻³ moles * (1L / 0.200moles) = 0.0195L.
The moles of HZ = Moles Z⁻ after the reaction and the volume is 0.010L + 0.0195L = 0.0295L. Thus, [Z⁻] is:
[Z⁻] = 3.9x10⁻³ moles / 0.0295L
[Z⁻] = 0.132M
Replacing in K expression:
4.17x10⁻⁹ = [OH⁻] [HZ] / [Z⁻]
4.17x10⁻⁹ = [X] [X] / [0.132]
5.51x10⁻¹⁰ = X²
X = 2.35x10⁻⁵M = [OH⁻]
As pOH = -log [OH⁻]
pOH = 4.63
And pH is
pH = 14 -pOH
pH = 9.37