A solution containing HCOOH and HCOO- in water is at a pH of 4.00. What can you conclude about the relative concentrations of HCOOH and HCOO- in this solution? The pKa for HCOOH is 3.75.

Respuesta :

Answer:

See explanation below.

Explanation:

To solve this problem we use the Henderson-Hasselbach formula:

pH = pKa + log [A⁻]/[HA] where [A⁻]/[HA]  is the ratio of the concentration of conjugate base to acid which is what we are seeking to answer, and we are given the values for pH and pKa.

pH = pKa + log [ HCOO⁻]/^[HCOOH]

4.00 = 3.75 + log [ HCOO⁻]/^[HCOOH]

4.00 - 3.75 = log [ HCOO⁻]/^[HCOOH]

0.25 =  log [ HCOO⁻]/^[HCOOH]

Taking antilog to both sides of the equation

10^0.25 =  [ HCOO⁻]/^[HCOOH]

1.77 =  [ HCOO⁻]/^[HCOOH]

This ratio tell us that the concentration of  conjugate base is 1.77 times the concentration of HCOOH, so we have a buffer solution   since HCOOH is a weak acid and therefore it will not dissociate into H⁺ and HCOO⁻ to an appreciable extent.

Remember buffer solutions are prepared by adding the salt of the conjugate base to the weak acid according to the desired pH.

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