A buffer solution is prepared by mixing 65.5 mL of 0.768 M sodium hydrogen sulfi

Question

A buffer solution is prepared by mixing 65.5 mL of 0.768 M sodium hydrogen sulfite with 25.3 mL of 0.0706 M sodium sulfite.
Calculate the pH (to two decimal places) of this solution. pKa hydrogen sulfite ion is 7.19

Assume the 5% approximation is valid and that the volumes are additive.

Calculate the pH (to two decimal places) of the buffer solution after the addition of 0.0173 g of sodium sulfite (Na2SO3) to the buffer solution above.
Assume 5% approximation is valid and that the volume of solution does not change.

Explanation / Answer

1) pH = pka + log [ conjugate base] /[acid]

acid is NaHSO3 ,   conjugate base is Na2SO3.

we find final concentrations of NaHSO3 and Na2SO3 by using M1V1 = M2V2 formula

V2 = final vol = 65.5+25.3 = 90.8 ml

now for NaHSO3 , 0.768 x 65.5 = M2 x 90.8 , M2 = 0.554   is final conc of NaHSO3

now for Na2SO3 , 0.0706 x 25.3 = M2 x 90.8 , M2 = 0.01967 is final conc of Na2SO3,

pH = pka + log [Na2SO3]/[NaHSO3]

pH = 7.19 + log ( 0.01967 /0.554) = 5.74

2)   Na2SO3 moles added = mass of Na2SO3 / Molar mass of Na2SO3 = 0.0173/126.04 = 0.00013725

Na2SO3 moles before addition = M x V ( in liters) = 0.01967 x 90.8/1000 = 0.001786

final Na2SO3 moles = 0.001786+0.00013725 = 0.001923

[Na2SO3] = ( 0.001923/0.0908) = 0.0212

pH = 7.19+ log ( 0.0212/0.554) = 5.77

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