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Assuming all the nh3 dissolves and that the volume of the solution remains at 0.300 l , calculate the ph of the resulting solution.

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Assume p=735 Torr V= 7.6L R=62.4 T= 295 PV-nRT (735 Torr)(7.60L)= n (62.4Torr-Litres/mole-K)(295K) 0.30346 moles of NH3 Find moles 0.300L solution of 0.300 M HCL = 0.120 moles of HCL 0.30346 moles of NH3 reacts with 0.120 moles of HCL producing 0.120 moles of NH4+ ION, and leaving 0.18346 mole sof NH3 behind Find molarity 0.120 moles of NH4+/0.300L = 0.400 M NH4+ 0.18346 moles of NH3/0.300L = 0.6115 M NH3 NH4OH --> NH4 & OH- Kb = [NH4+][OH]/[NH4OH] 1.8 e-5=[0.300][OH-]/[0.6115] [OH-]=1.6e-5 pOH= 4.79 PH=9.21 .

The pH of the solution of dissolve ammonia has been calculated as 9.21.

The volume of the solution has been 0.3 L.

Applying the ideal gas equation,

PV =nRT

n will be the moles of ammonia in the solution

735 torr [tex]\times[/tex] 0.3 L = n [tex]\times[/tex] 62.4 [tex]\times[/tex] 295 K

n = 0.01 moles.

Since the remaining solution is HCl, the moles of HCl will be: 2 times of volume

Moles of HCl = 0.3 [tex]\times[/tex] 0.3 moles

Moles of HCl = 0.09 moles

0.01 moles of ammonia has remained in the sample.

Molarity = moles per liter

Molarity = [tex]\rm \dfrac{0.1}{0.3}[/tex]

Molarity = 0.33 M

The pH can be calculated as:

[tex]\rm K_b[/tex] = 1.6

pOH = 4.79

pH = 14 - pOH

pH = 14 - 4.79

pH = 9.21

The pH of the solution of dissolved ammonia has been calculated as 9.21.

For more information about pH, refer to the link:

https://brainly.com/question/491373

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