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laiz [17]
3 years ago
5

Calculate the change in ph when 3.00 ml of 0.100 m hcl(aq) is added to 100.0 ml of a buffer solution that is 0.100 m in nh3(aq)

and 0.100 m in nh4cl(aq). a list of ionization constants can be found here.
Chemistry
1 answer:
nikklg [1K]3 years ago
7 0

The change in pH is calculated by:

pOH = Protein kinase B + log [NH4+]/ [NH3] 

Protein kinase B of ammonia = 4.74 

initial potential of oxygen hydroxide= 4.74 + log 0.100/0.100 = 4.74 
pH = 14 - 4.74=9.26 

moles NH4+ = moles NH3 = 0.100 L x 0.100 M = 0.0100 
moles H+ added = 3.00 x 10^-3 L x 0.100 M=0.000300 

NH3 + H+ = NH4+ 
moles NH3 = 0.0100 - 0.000300=0.00970 
moles NH4+ = 0.0100 + 0.000300=0.0103 

pOH = 4.74 + log 0.0103/ 0.00970= 4.77 
oH = 14 - 4.77 = 9.23 

the change is  = 9.26 - 9.23 =0.03 

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<u>Answer:</u> The freezing point of solution is -0.454°C

<u>Explanation:</u>

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\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

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W_{solvent} = Mass of solvent (water) = 550.0 g

Putting values in above equation, we get:

0-\text{Freezing point of solution}=2\times 1.86^oC/m\times \frac{5\times 1000}{74.55g/mol\times 550}\\\\\text{Freezing point of solution}=-0.454^oC

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