For 250.0 mL of a buffer solution that is 0.295 M in CH3CH2NH2 and 0.275 M...

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For 250.0 mL of a buffer solution that is 0.295 M in CH3CH2NH2 and 0.275 M...

For 250.0 mL of a buffer solution that is 0.295 M in CH3CH2NH2 and 0.275 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.010 mol of NaOH.

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Answer 1

Answer #1

Volume of buffer solution = 250ML = 0.25L Moles of CH₃ CH₃NH = 0.25L x 0.295 mol/l = 0.07375 mol CH₃ CH₃NH₂ Moles of CH₂CH₃NH [OH-] = 0.010 ml NaOH x 1 mol OH 1 mol NaoH = 0.010 mol OH C₂H NH3 + oH H₂O + CH ₂NH₂ Initial 0.06875 0.010 0.07375 Final -0.

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Answer 2

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