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marissa [1.9K]
4 years ago
11

A 75.0-ml volume of 0.200M NH3 (Kb=1.8 x 10^-5) is titrated with 0.500M NHO3. Calculate the pH after the addition of 13.0mL of N

H3.
Chemistry
1 answer:
poizon [28]4 years ago
3 0

The solution would be like this for this specific problem:

<span>moles NH3 = 0.0750 L x 0.200 M=0.015 </span><span>
<span>moles HNO3 = 0.0130 L x 0.500 M=0.0065 </span>

<span>NH3 + H+ = NH4+ </span>
<span>moles NH3 = 0.015 - 0.0065=0.0085 </span>
<span>moles NH4+ = 0.0065 </span>
<span>pKb = 4.74 </span>
<span>pOH = 4.74 + log 0.0065/0.0085=4.62 </span>
pH = 9.38</span>

So the pH after the addition of 13.0 mL of HNO_3 is 9.38.

I am hoping that these answers have satisfied your query and it will be able to help you in your endeavors, and if you would like, feel free to ask another question.

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