Calculate the pH of a buffer solution after the addition of 20.0 mL of 0.200 M...

Calculate the pH of a buffer solution after the addition of 20.0 mL of 0.200 M NaOH to 80.0 mL of 0.0500 M HC3H5O2 and 0.0250 M NaC3H5O2. (Ka = 1.4 x 10^-5)

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Expert Solution

Moles of HC3H5O2 = M x V = 0.05 x 80/1000 = 0.004

moles of NaC3H5O2 = M x V = 0.025 x 80/1000 = 0.002

NaOH moles added = Mx V = 0.2 x 20/1000 = 0.004

now NaOH moles = HC3H5O2

NaOH reacts with HC3H5O2 and converts it to NaC3H5O2

thus all HC3H5O2 gets converted to NaC3H5O2

thus NaC3H5O2 moles =0.004+0.002 = 0.006, solution vol = 100 ml = 0.1L

now we have equilibrium

C3H5O2-(aq) + H2O (l) <---> HC3H5O2 (aq) + OH-(aq) , Kb = Kw/Ka = 10^ -14/1.4x10^-5 = 7.14 x 10^-10

Intially C3H5O2- moles = 0.006 , HC3H5O2 moles = OH- moles = 0

at equilibrium C3H5O2- moles = 0.006-X , HC3H5O2 moles = OH- moles = X

[C3H5O2-] = ( 0.006-X) /0.1 , [HC3H5O2] = [OH-] = X/0.1

Kb = [HC3H5O2] [OH-] / [C3H5O2-]

7.14 x 10^ -10 = ( X/0.1) ( X/0.1) / (0.006-X) /0.1

7.14 x 10^ -11 = X^2 / ( 0.006-X) ( here 0.006-X is nearly 0.006 since we get very less value of X as Kb is very low)

hence 7.14 x 10^ -11 = X^2 / ( 0.006)

X = OH- moles = 6.5465 x 10^ -7

[OH-] = ( 6.5465x10^ -7) /0.1 = 6.5465 x 10^ -6 M

pOH = -log [OH-] = -log ( 6.5465 x 10^ -6) = 5.2

pH = 14-pOH = 14-5.2 = 8.82

queen_honey_blossom answered 2 hours ago

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