A student prepares a lactic acid-sodium lactate buffer solution by mixing 40.mL of 0.50MHC3H5O3(aq) with 200.mL of 1.0M NaC3H5O3(aq) . The pKa of HC3H5O3 is 3.08. What is the pH of the resulting solution?

pH=3.38

Explanation:

We are given the concentrations of the weak acid and its conjugate base so we need to use the henderson hasslebalch equation

pH=pKa + log (concentration A-/concentration of HA)

A- : conjugate base NaC3H5O3

HA: weak acid HC3H5O3

so if we plug in what we are given

pH= 3.08 + log (1M/.5M)

pH= 3.38

i think its 100.

hope this !

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