Respuesta :
Answer:
Following are the answer to this question:
Explanation:
The value of pH solution is =5.17 So, the p^{OH}:
[tex]p^{OH}[/tex]=14-56.17
=8.823
The volume of the [tex]NH_{3}[/tex] = 40.00 ml
convert into the liter= 0.040L
The value of the concentrated [tex]NH_{3}[/tex] =0.10 M
The volume of the [tex]NH_{4}Cl[/tex]= 50.00 ml
convert into the liter= 0.050L
The value of concentrated [tex]NH_{4}Cl[/tex]= 0.10 M
The volume of the [tex]H_{2}So_{4}[/tex]= 30 ml
convert into the liter= 0.030L
The value of concentrated [tex]H_2So_4[/tex]=0.05 M
Calculating total volume=(0.40+0.050+0.030)
=0.120 L
calculating the new concentrated value of [tex]NH_3[/tex] = [tex]\frac{0.10\times 0.040}{0.120}= 0.33 \ M[/tex]
calculating the new concentrated value of [tex]NH_4Cl[/tex]= [tex]\frac{0.050\times 0.10}{0.120}= 0.04166 \ M[/tex]calculating the new concentrated value of [tex]H_2So_4= \frac{0.030\times 0.05}{0.120}= 0.0125 \ M[/tex] when 1 mol [tex]H_2So_4[/tex] produced 2 mols [tex]H^{+}[/tex] so, 0.0125 in [tex]H_2So_4[/tex]produced:
[tex]=4 \times (2 \times 0.0125) \ mol H^{+}\\\\= 0.025 mol H^{+}[/tex]
create the ICE table:
[tex]NH_3 \ \ \ \ \ \ \ \ + H^{+} \ \ \ \ \ \ \longrightarrow NH_4^{+}[/tex]
I (m) 0.033(m) 0.025 0.04166
C -0.025 -0.025 + 0.025
E 8.3\times 10^{-3} 0 0.0667
now calculating pH:
when ph= 8.83:
[tex]P^{H}= p^{kb}|+ \log\frac{[NH_4^{+}]}{[NH_3]}\\\\8.83=p^{kb}+\log\frac{0.0667}{8.3 \times 10^{-3}}\\\\p^{kb}=8.83-0.9069\\\\ \ \ \ =7.7231 \\\\\ The P^{kb} \ for \ NH_3 \ is =7.7231\\\\\ The P^{kb} \ for N^{+}H_4=14-7.7231\\\\\ \ \ \ \ \ =6.2769[/tex]
