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

Following are the answer to this question:

Explanation:

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

=14-56.17

      =8.823

The volume of the = 40.00 ml  

convert into the liter= 0.040L

The value of the concentrated =0.10 M

The volume of the = 50.00 ml

convert into the liter= 0.050L

 The value of concentrated = 0.10 M

The volume of the = 30 ml

convert into the liter= 0.030L  

The value of concentrated =0.05 M

Calculating total volume=(0.40+0.050+0.030)

                                       =0.120 L

calculating the new concentrated value of =

calculating the new concentrated value of = calculating the new concentrated value of when 1 mol produced 2 mols so, 0.0125 in produced:

create the ICE table:    

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:

possibly, yes.

explanation:

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