# What is the K_a for "HNO"_2 if its starting concentration of "0.14 M" decreased to "0.11 M" by the time the reaction species reached equilibrium?

Jun 19, 2017

${K}_{\textrm{a}} = 0.008$.

#### Explanation:

Let's pretend we are trying to solve this question using an ICE table. We would write

$\textcolor{w h i t e}{m m m m m m} \text{HNO"_2 + "H"_2"O" ⇌ "H"_3"O"^"+" + "NO"_2^"-}$
$\text{I/mol·L"^"-1} : \textcolor{w h i t e}{m} 0.14 \textcolor{w h i t e}{m m m m m m m l} 0 \textcolor{w h i t e}{m m m l l} 0$
$\text{C/mol·L"^"-1":color(white)(m)"-"xcolor(white)(mmmmmmml)"+"xcolor(white)(mmm)"+} x$
$\text{E/mol·L"^"-1":color(white)(l) "0.14-} x \textcolor{w h i t e}{m m m m m m l l} x \textcolor{w h i t e}{m m m l l} x$

In this case, we are told that the equilibrium concentration of the acid, i.e., ["HNO"_2] = "0.11 mol/L".

In other words,

$0.14 - x = 0.11$

$x = 0.14 - 0.11 = 0.03$

If we insert this number into the ICE table, we get

$\textcolor{w h i t e}{m m m m m m} \text{HNO"_2 + "H"_2"O" ⇌ "H"_3"O"^"+" + "NO"_2^"-}$
$\text{I/mol·L"^"-1} : \textcolor{w h i t e}{m} 0.14 \textcolor{w h i t e}{m m m m m m m l} 0 \textcolor{w h i t e}{m m m l l} 0$
$\text{C/mol·L"^"-1":color(white)(l)"-0.03"color(white)(mmmmmm)"+0.03"color(white)(ml)"+0.03}$
$\text{E/mol·L"^"-1} : \textcolor{w h i t e}{l l} 0.11 \textcolor{w h i t e}{m m m m m m m} 0.03 \textcolor{w h i t e}{m m l} 0.03$

We can now insert these numbers into the ${K}_{\textrm{a}}$ expression.

K_text(a) = (["H"_3"O"^"+"]["NO"_2^"-"])/(["HNO"_2]) = (0.03 × 0.03)/0.11 = 0.008