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

Answer:

Following are the answer to this question:

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 $NH_3$ = $\frac{0.10\times 0.040}{0.120}= 0.33 \ M$

calculating the new concentrated value of $NH_4Cl$= $\frac{0.050\times 0.10}{0.120}= 0.04166 \ M$calculating the new concentrated value of $H_2So_4= \frac{0.030\times 0.05}{0.120}= 0.0125 \ M$ when 1 mol $H_2So_4$ produced 2 mols $H^{+}$ so, 0.0125 in $H_2So_4$produced:

$=4 \times (2 \times 0.0125) \ mol H^{+}\\\\= 0.025 mol H^{+}$

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:

$P^{H}= p^{kb}|+ \log\frac{[NH_4^{+}]}{[NH_3]}\\\\8.83=p^{kb}+\log\frac{0.0667}{8.3 \times 10^{-3}}\\\\p^{kb}=8.83-0.9069\\\\ \ \ \ =7.7231 \\\\\ The P^{kb} \ for \ NH_3 \ is =7.7231\\\\\ The P^{kb} \ for N^{+}H_4=14-7.7231\\\\\ \ \ \ \ \ =6.2769$