Question

The pH of a solution prepared by mixing 40.00 mL of 0.10 M NH3 with 50.00 mL of 0.10 M NH4Cl and 30mL of 0.05 M H2SO4 is 5.17. Assume that the volume of the solutions are additive . What would be the Ka for NH4

Answer 1

Following are the answer to this question:

Step-by-step explanation:

The value of pH solution is =5.17 So, the p^{OH}:

`p^(OH)`=14-56.17

=8.823

The volume of the
`NH_(3)` = 40.00 ml

convert into the liter= 0.040L

The value of the concentrated
`NH_(3)` =0.10 M

The volume of the
`NH_(4)Cl`= 50.00 ml

convert into the liter= 0.050L

The value of concentrated
`NH_(4)Cl`= 0.10 M

The volume of the
`H_(2)So_(4)`= 30 ml

convert into the liter= 0.030L

The value of concentrated
`H_2So_4`=0.05 M

Calculating total volume=(0.40+0.050+0.030)

=0.120 L

calculating the new concentrated value of
`NH_3` =
`(0.10* 0.040)/(0.120)= 0.33 \ M`

calculating the new concentrated value of
`NH_4Cl`=
`(0.050* 0.10)/(0.120)= 0.04166 \ M`calculating the new concentrated value of
`H_2So_4= (0.030* 0.05)/(0.120)= 0.0125 \ M` when 1 mol
`H_2So_4` produced 2 mols
`H^(+)` so, 0.0125 in
`H_2So_4`produced:

`=4 * (2 * 0.0125) \ mol H^(+)\\\\= 0.025 mol H^(+)`

create the ICE table:

`NH_3 \ \ \ \ \ \ \ \ + H^(+) \ \ \ \ \ \ \longrightarrow NH_4^(+)`

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:

`P^(H)= p^(kb)|+ \log([NH_4^(+)])/([NH_3])\\\\8.83=p^(kb)+\log(0.0667)/(8.3 * 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`