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tigry1 [53]
3 years ago
5

Calculate the ph at the equivalence point for the titration of 0.190 m methylamine (ch3nh2) with 0.190 m hcl. the kb of methylam

ine is 5.0× 10–4.
Chemistry
1 answer:
bija089 [108]3 years ago
3 0
<span>HCl(aq)+CH3NO2(aq)⇄CH3NH3+(aq)+CL-(aq) Given that the concentrations of the HCl and methyamine are equal, so at the equivalence point the volume is doubled. The resulting concentration of CH3NH3+ is 0.095M CH3NH3+(aq)⇄H+(aq)+CH3NH2(aq) I 0.095 0 0 C -x +x +x E 0.095-x x x Ka=x^2/0.095-x Ka=1.0x10^-14/5.0x10^-4 =2.0x10^-11 x^2/0.095-x=2.0x10^-11 x=1.38x10^-6 pH=-log(1.38x10^-6) =5.9</span>
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Answer:

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(ii) 6(-1/√2)² - 3

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