# How would you use the Henderson-Hasselbalch equation to calculate the pH of a buffer solution that is 0.27 M in formic acid (HCO2H) and 0.50 M in sodium formate (HCO2Na)?

Nov 17, 2015

$\text{pH} = 4.02$

#### Explanation:

Your buffer solution contains formic acid, $\text{HCOOH}$, a weak acid, and sodium formate, $\text{HCOONa}$, the salt of its conjugate base, the formate anion, ${\text{HCOO}}^{-}$.

The Henderson - Hasselbalch equation allows you to calculate the pH of the buffer by using the $p {K}_{a}$ of the weak acid and the ratio that exists between the concentrations of the weak cid and conjugate base.

color(blue)("pH" = pK_a + log( (["conjugate base"])/(["weak acid"])))

The $p {K}_{a}$ of formic acid is equal to $3.75$

http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf

Before plugging in the values given to you, try to predict what you expect the pH of the solution to be. Notice that at equal concentrations of weak acid and conjugate base, the log term is equal to zero.

In this case, the pH of the solution will be equal to the acid's $p {K}_{a}$. Now, if you have more conjugate base than weak acid, like you have here, you would expect the log term to return a positive value.

This means that the pH will actually increase, which is what you should expect to see in this case.

"pH" = pK_a + log( (["HCOO"^(-)])/(["HCOOH"]))

"pH" = 3,.75 + log( (0.50color(red)(cancel(color(black)("M"))))/(0.27color(red)(cancel(color(black)("M")))))

$\text{pH} = 3.75 + 0.268 = \textcolor{g r e e n}{4.02}$

Indeed, the pH of the buffer is higher than the $p {K}_{a}$ of the acid.