Let's pretend we are trying to solve this question using an ICE table. We would write
#color(white)(mmmmmm)"HNO"_2 + "H"_2"O" ⇌ "H"_3"O"^"+" + "NO"_2^"-"#
#"I/mol·L"^"-1":color(white)(m)0.14color(white)(mmmmmmml)0color(white)(mmmll)0 #
#"C/mol·L"^"-1":color(white)(m)"-"xcolor(white)(mmmmmmml)"+"xcolor(white)(mmm)"+"x #
#"E/mol·L"^"-1":color(white)(l) "0.14-"xcolor(white)(mmmmmmll)xcolor(white)(mmmll)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
#color(white)(mmmmmm)"HNO"_2 + "H"_2"O" ⇌ "H"_3"O"^"+" + "NO"_2^"-"#
#"I/mol·L"^"-1":color(white)(m)0.14color(white)(mmmmmmml)0color(white)(mmmll)0 #
#"C/mol·L"^"-1":color(white)(l)"-0.03"color(white)(mmmmmm)"+0.03"color(white)(ml)"+0.03"#
#"E/mol·L"^"-1":color(white)(ll) 0.11color(white)(mmmmmmm)0.03color(white)(mml)0.03#
We can now insert these numbers into the #K_text(a)# expression.
#K_text(a) = (["H"_3"O"^"+"]["NO"_2^"-"])/(["HNO"_2]) = (0.03 × 0.03)/0.11 = 0.008#
