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What is the ph of a 2.00 m hno2 solution if ka=5.00 e-6?

User Doobean
by
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1 Answer

6 votes

Answer: -

2.5

Explanation: -

Initial [HNO₂] = 2.00 M

ka = 5.00 x 10 ⁻⁶

ICE table for HNO₂ =

[
\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]

Ka =
(x2)/(2.00-x)

5.00 x 10-6 =
(x2)/(2.00-x)

x = 0.0032 M

pH = - log [H⁺]

= - log 0.0032

= 2.5

User Marangely
by
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