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For 260.0 mL of a buffer solution that is 0.2869 M in CH3CH2NH2 and 0.2669 M...

For 260.0 mL of a buffer solution that is 0.2869 M in CH3CH2NH2 and 0.2669 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.0200 mol of NaOH(Kb=5.6⋅10−4).

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Expert Solution

Ans - Since this is a buffer solution, we can use the Henderson Hasselbalch eqn. to calculate the pH
pKa = 10.636 for CH3CH2NH3Cl
pH = pKa + log([CH3CH2NH2] / [CH3CH2NH3Cl]) = 10.636 + Log ( 0.2869 / 0.2669 ) = 10.667

Final pH:
moles CH3CH2NH2 = 0.260 L x 0.2869 = 0.0746 M

moles CH3CH2NH3Cl = 0.260 x .2669 = .0693 M

CH3CH2NH3+ + OH- = CH3CH2NH2 + H2O

moles CH3CH2NH3+ = .0693 - .02 = .0493

moles CH3CH2NH2 = 0.0746 + .02 = .0946

Can use the Henderson Hasselbalch eqn. since this is still a healthy buffer (still decent amount of base and acid present).

[CH3CH2NH3+] = 0.0493 / .260 = 0.189 M

[CH3CH2NH2] = .0946 / .260 = 0.363 M

pH = 10.636 + log(.363 / 0.189) = 10.91

queen_honey_blossom answered 1 year ago
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