# What is the pH of a .200 M solution of formic acid?

Oct 18, 2016

#### Answer:

And $p {K}_{a}$ for formic acid is?

$p H = 2.23$

#### Explanation:

$H C \left(= O\right) O H + {H}_{2} O r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + H C \left(= O\right) {O}^{-}$

${K}_{a} = \frac{\left[{H}_{3} {O}^{+}\right] \left[H C {O}_{2}^{-}\right]}{\left[H C \left(= O\right) O H\right]}$

$=$ ${x}_{1}^{2} / \left(0.200 - {x}_{1}\right)$ $=$ $1.77 \times {10}^{-} 4$.

Of course I had to look up ${K}_{a} \text{ formic acid}$ on the interwebz.

If, ${x}_{1} < 0.200$, then $\left(0.200 - {x}_{1}\right) \cong 0.200$

Thus ${x}_{1}^{2}$ $=$ $1.77 \times {10}^{-} 4 \times \left(0.200\right)$

${x}_{1}$ $=$ $\sqrt{1.77 \times {10}^{-} 4 \times \left(0.200\right)}$$=$ $5.95 \times {10}^{-} 3$

${x}_{2}$ $=$ $\sqrt{1.77 \times {10}^{-} 4 \times \left(0.200 - 5.95 \times {10}^{-} 3\right)}$$=$ $5.86 \times {10}^{-} 3$

${x}_{3}$ $=$ $\sqrt{1.77 \times {10}^{-} 4 \times \left(0.200 - 5.86 \times {10}^{-} 3\right)}$$=$ $5.86 \times {10}^{-} 3$

So successive approximations, leads us to $\left[{H}_{3} {O}^{+}\right] = 5.86 \times {10}^{-} 3 \cdot m o l \cdot {L}^{-} 1$.

$p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right] = - {\log}_{10} \left\{5.86 \times {10}^{-} 3\right\}$ $=$ $2.23$

As carboxylic acids go, this is a fairly low $p H$. Why should formic acid be more acidic than say propionic acid?