What is the pH of the solution?

Question

Buffers resist the change in pH when small amounts of either acid or base are added. Buffers are often made from an acid and a salt of that same acid such as with methanoic acid (formic acid) and sodium formate.

If a 0.016 M solution of each were used and K_a for formic acid is 1.8 x 10^-4, calculate the pH of the solution. The equilibrium reaction is as shown below.                                                                                                                                                                          

HCOOH _(aq) + H₂O _(l) ⇋ H₃O⁺ _(aq) + HCOO^- _(aq)

Answer 1

Buffers resist the change in pH when small amounts of either acid or base are added. Buffers are often made from an acid and a salt of that same acid such as with methanoic acid (formic acid) and sodium formate.

If a 0.016 M solution of each were used and K_a for formic acid is 1.8 x 10^-4, calculate the pH of the solution. The equilibrium reaction is as shown below.                                                                                                                                                                         

HCOOH _(aq) + H₂O _(l) ⇋ H₃O⁺ _(aq) + HCOO^- _(aq)

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You have to use the Hendersen-Hasselbach Equation.

pH = pK_a + log([A^–]/[HA])

for Formic acid K_a = 1.8 * 10^-4 thus the pKa = -log(1.8 * 10^-4) ~= 3.75

[A-] is 0.016M Formate

[HA] is 0.016M Formic acid

(Note the concentration of formate and formic acid are the same. What happens to the ratio of A-/HA and the log term?)

Plug them all in to the equation.

pH = 3.75 + log ((0.016)/(0.016))

= 3.75 + log (1)

= 3.75 + 0

= 3.75