Answer: -
2.5
Explanation: -
Initial [HNO₂] = 2.00 M
ka = 5.00 x 10 ⁻⁶
ICE table for HNO₂ =
[[tex] \left[\begin{array}{cccc}concentrations&HNO2&H+&NO2\\Initial&2.00&0&0\\Change&-x&+x&+x\\Equilibrium&2.00-x&x&x\end{array}\right] [/tex]
Ka = [tex] \frac{x2}{2.00-x} [/tex]
5.00 x 10-6 = [tex] \frac{x2}{2.00-x} [/tex]
x = 0.0032 M
pH = - log [H⁺]
= - log 0.0032
= 2.5
