Respuesta :
Answer:
C
Explanation:
The question asks to calculate the pH at equivalence point of the titration between ammonia and hydrochloric acid
Firstly, we write the equation of reaction between ammonia and hydrochloric acid.
NH3(aq)+HCl(aq)→NH4Cl(aq)
Ionically:
HCl + NH3 ---> NH4 + Cl-
Firstly, we calculate the number of moles of the ammonia as follows:
from c = n/v and thus, n = cv = 0.36 × 1 = 0.36 moles
At the equivalence point, there is equal number of moles of ammonia and HCl.
Hence, volume of HCl = number of moles/molarity of HCl = 0.36/0.74 = 0.486L
Hence, the total volume of solution will be 1 + 0.486 = 1.486L
Now, we calculate the concentration of the ammonium ions = 0.36/1.486 = 0.242M
An ICE TABLE IS USED TO FIND THE CONCENTRATION OF THE HYDROXONIUM ION(H3O+). ICE STANDS FOR INITIAL, CHANGE AND EQUILIBRIUM.
NH4+ H2O ⇄ NH3 H3O+
I 0.242 0 0
C -X +x +X
E 0.242-X X X
Since the question provides us with the base dissociation constant value K b, we can calculate the acid dissociation constant value Ka
To find this, we use the mathematical equation below
K a ⋅ K b = K w
, where K w- the self-ionization constant of water, equal to
10 ^-14 at room temperature
This means that you have
K a = K w.K b = 10 ^− 14 /1.8 * 10^-5 = 5.56 * 10^-10
Ka = [NH3][H3O+]/[NH4+]
= x * x/(0.242-x)
Since the value of Ka is small, we can say that 0.242-x ≈ 0.242
Hence, K a = x^2/0.242 = 5.56 * 10^-10
x^2 = 0.242 * 5.56 * 10^-10 = 1.35 * 10^-10
x = 0.00001161895
[H3O+] = 0.00001161895
pH = -log[H3O+]
pH = -log[0.00001161895 ] = 4.94
The pH at the equivalence point of the titration is:
C) 4.94
Chemical equation:
NH₃(aq)+HCl(aq)→NH₄Cl(aq)
Ionic chemical equation:
HCl + NH₃ ---> [tex]NH_4^+ + Cl^-[/tex]
Calculation for number of moles:
c = n/v
n = cv = 0.36 × 1 = 0.36 moles
At the equivalence point, there is equal number of moles of ammonia and HCl.
Hence, volume of HCl = number of moles/molarity of HCl = 0.36/0.74 = 0.486L
Hence, the total volume of solution will be 1 + 0.486 = 1.486L
Calculation for concentration of the ammonium ions = 0.36/1.486 = 0.242M
[tex]NH_4^+[/tex] H₂O ⇄ NH₃ [tex]H_3O^+[/tex]
I 0.242 0 0
C -X +x +X
E 0.242-X X X
Since the question provides us with the base dissociation constant value K b, we can calculate the acid dissociation constant value :Ka
To find this, we use the mathematical equation below:
K a ⋅ K b = K w
Kw = [tex]10^{-14}[/tex]
[tex]K_a = K_w*K_b\\\\ 10^{− 14} /1.8 * 10^{-5} = 5.56 * 10^{-10}\\\\K_a = [NH_3][H_3O^+]/[NH_4^+]\\\\K_a = x * x/(0.242-x)[/tex]
Since, the value of Ka is small, we can say that 0.242-x ≈ 0.242
[tex]K_a = x^2/0.242 = 5.56 * 10^{-10}\\\\x^2 = 0.242 * 5.56 * 10^{-10} = 1.35 * 10^{-10}\\\\x = 0.00001161895\\\\H_3O^+= 0.00001161895\\\\pH = -log[H_3O^+]\\\\pH = -log[0.00001161895 ] = 4.94[/tex]
Thus, correct option is C. The pH at the equivalence point of the titration is: 4.94
Find more information about Dissociation constant here:
brainly.com/question/3006391
