Respuesta :
Answer:
Explanation:
Given that:
Concentration of [tex]HC_2H_3O_2 \ (M_1)[/tex] = 0.105 M
Volume of [tex]HC_2H_3O_2 \ (V_1)[/tex] = 20.0 mL
Concentration of [tex]NaOH (M_2)[/tex] = 0.125 M
The chemical reaction can be expressed as:
[tex]HC_2H_3O_2_{(aq)} + NaOH _{(aq)} \to NaC_2H_3O_2_{(aq)} + H_2O_{(l)}[/tex]
Using the ICE Table to determine the equilibrium concentrations.
[tex]HC_2 H_3 O_2 _{(aq)} + H_2O _{(l) } \to C_2 H_3O_2^- _{(aq)} + H_3O^+_{ (aq)}[/tex]
I 0.105 0 0
C -x +x +x
E 0.105 - x x x
[tex]K_a = \dfrac{[C_2H_5O^-_2][H_3O^+]}{[HC_2H_3O_2]}[/tex]
[tex]K_a = \dfrac{(x)(x)}{(0.105-x)}[/tex]
Recall that the ka for [tex]HC_2H_3O_2= 1.8 \times 10^{-5}[/tex]
Then;
[tex]1.8 \times 10^{-5} = \dfrac{(x)(x)}{(0.105 -x)}[/tex]
[tex]1.8 \times 10^{-5} = \dfrac{x^2}{(0.105 -x)}[/tex]
By solving the above mathematical expression;
x = 0.00137 M
[tex]H_3O^+ = x = 0.00137 \ M \\ \\ pH = - log [H_3O^+] \\ \\ pH = - log ( 0.00137 )[/tex]
pH = 2.86
Hence, the initial pH = 2.86
b) To determine the volume of the added base needed to reach the equivalence point by using the formula:
[tex]M_1 V_1 = M_2 V_2[/tex]
[tex]V_2= \dfrac{M_1V_1}{M_2}[/tex]
[tex]V_2= \dfrac{0.105 \ M \times 20.0 \ mL }{0.125 \ M}[/tex]
[tex]V_2 = 16.8 mL[/tex]
Thus, the volume of the added base needed to reach the equivalence point = 16.8 mL
c) when pH of 5.0 mL of the base is added.
The Initial moles of [tex]HC_2H_3O_2 =[/tex] molarity × volume
[tex]= 0.105 \ M \times 20.0 \times 10^{-3} \ L[/tex]
[tex]= 2.1 \times 10^{-3}[/tex]
number of moles of 5.0 NaOH = molarity × volume
number of moles of 5.0 NaOH = [tex]0.625 \times 10^{-3}[/tex]
After reacting with 5.0 mL NaOH, the number of moles is as follows:
[tex]HC_2 H_3 O_2 _{(aq)} + NaOH _{(aq)} \to NaC_2H_3O_2_{(aq)} + H_2O{ (l)}[/tex]
Initial moles [tex]2.1*10^{-3}[/tex] [tex]0.625 * 10^{-3}[/tex] 0 0
F(moles) [tex](2.1*10^{-3} - 0.625 \times 10^{-3})[/tex] 0 [tex]0.625 \times 10^{-3}[/tex] [tex]0.625 \times 10^{-3}[/tex]
The pH of the solution is then calculated as follows:
[tex]pH = pKa + log \dfrac{[base]} {[acid]}[/tex]
Recall that:
pKa for [tex]HC_2H_3O_2=4.74[/tex]
Then; we replace the concentration with the number of moles since the volume of acid and base are equal
∴
[tex]pH = 4.74 + log \dfrac{0.625 \times 10^{-3}}{1.475 \times 10^{-3}}[/tex]
pH = 4.37
Thus, the pH of the solution after the addition of 5.0 mL of NaOH = 4.37
d)
We need to understand that the pH at 1/2 of the equivalence point is equal to the concentration of the base and the acid.
Therefore;
pH = pKa = 4.74
e) pH at the equivalence point.
Here, the pH of the solution is the result of the reaction in the [tex](C_2H_3O^-_2)[/tex] with [tex]H_2O[/tex]
The total volume(V) of the solution = V(acid) + V(of the base added to reach equivalence point)
The total volume(V) of the solution = 20.0 mL + 16.8 mL
The total volume(V) of the solution = 36.8 mL
Concentration of [tex](C_2H_3O^-_2)[/tex] = moles/volume
= [tex]\dfrac{2.1 \times 10^{-3} \ moles}{0.0368 \ L}[/tex]
= 0.0571 M
Now, using the ICE table to determine the concentration of [tex]H_3O^+[/tex];
[tex]C_2H_5O^-_2 _{(aq)} + H_2O_{(l)} \to HC_2H_3O_2_{(aq)} + OH^-_{(aq)}[/tex]
I 0.0571 0 0
C -x +x +x
E 0.0571 - x x x
Recall that the Ka for [tex]HC_2H_3O_2[/tex] = [tex]1.8 \times 10^{-5}[/tex]
[tex]K_b = \dfrac{K_w}{K_a} = \dfrac{1.0\times 10^{-14}}{1.8 \times 10^{-5} } \\ \\ K_b = 5.6 \times 10^{-10}[/tex]
[tex]k_b = \dfrac{[ HC_2H_3O_2] [OH^-]}{[C_2H_3O^-_2]}[/tex]
[tex]5.6 \times 10^{-10} = \dfrac{x *x }{0.0571 -x}[/tex]
[tex]x = [OH^-] = 5.6 \times 10^{-6} \ M[/tex]
[tex][H_3O^+] = \dfrac{1.0 \times 10^{-14} }{5.6 \times 10^{-6} }[/tex]
[tex][H_3O^+] =1.77 \times 10^{-9}[/tex]
[tex]pH =-log [H_3O^+] \\ \\ pH =-log (1.77 \times 10^{-9}) \\ \\ \mathbf{pH = 8.75 }[/tex]
Hence, the pH of the solution at equivalence point = 8.75
f) The pH after 5.09 mL base is added beyond (E) point.
[tex]HC_2 H_3 O_2 _{(aq)} + NaOH _{(aq)} \to NaC_2H_3O_2_{(aq)} + H_2O{ (l)}[/tex]
Before 0.0021 0.002725 0
After 0 0.000625 0.0021
[tex][OH^-] = \dfrac{0.000625 \ moles}{(0.02 + 0.0218 ) \ L}[/tex]
[tex][OH^-] = \dfrac{0.000625 \ moles}{0.0418 \ L}[/tex]
[tex][OH^-] = 0.0149 \ M[/tex]
From above; we can determine the concentration of [tex]H_3O^+[/tex] by using the following method:
[tex][H_3O^+] = \dfrac{1.0 \times 10^{-14} }{0.0149}[/tex]
[tex][H_3O^+] = 6.7 \times 10^{-13}[/tex]
[tex]pH = - log [H_3O^+][/tex]
[tex]pH = -log (6.7 \times 10^{-13} )[/tex]
pH = 12.17
Finally, the pH of the solution after adding 5.0 mL of NaOH beyond (E) point = 12.17
