Question

7)A.)A buffer solution is 0.319 M in HCN and 0.347 M in KCN. If Ka for HCN is 4.0×10-10, what is the pH of this buffer solution?__

B.)A buffer solution is 0.479 M in H₂C₂O₄ and 0.292 M in KHC₂O₄. If K_a for H₂C₂O₄ is 5.9×10^-2, what is the pH of this buffer solution?

C.)A buffer solution is 0.353 M in NaHSO₃ and 0.366 M in Na₂SO₃. If K_a for HSO₃^- is 6.4×10^-8, what is the pH of this buffer solution?

Answer 1

Q7.

if a buffer use Hendesron hasselbach equation

pH = pKa + log(conjugate base / weak acid)

pKa = -log(Ka) = -log(4*10^-10) = 9.397940

now, HCN = 0.319, CN- = 0.347

substitute data

pH = 9.397940 + log(0.347 / 0.319 )

pH = 9.43447

b)

similar as before:

weak acid = H2C2O4

conjugate base = HC2O4 -

pKa = -log(Ka) = -log(5.9*10^-2) = 1.2291

pH = pKa + log(HC2O4- H2C24)

pH = 1.2291 + log(0.292 / 0.479)

pH 1.01414

c)

finally...

NaHSO3 --> Na+ + HSO3-

Na2SO3 --> 2Na+ SO3-2

this is the 2nd ionization of H2SO3 acid, so we need HSO3- Ka:

pKa = -log(Ka) = -log(6.4*10^-8) = 7.19382

then

pH= pKa+ log(SO3-2 / HSO3-)

pH = 7.19382 + log(0.366/ 0.353)

pH = 7.213233