# Problem: Knowing that NaNO2 is the salt of a strong base and weak acid calculate : a: pH of 0.100 M NaNO2 solution. Ka of HNO2 = 4.6 x 10-4.b: pH of the solution obtained on adding 20.0 ml of a 0.100 M HCl solution to 60.0 ml of a 0.100 M NaNO2 .c: pH of the solution obtained on adding 120.0 ml of a 0.100 M HCl solution to 60.0 ml of a 0.100 M NaNO2

###### FREE Expert Solution

To calculate pH of starting NaNO2 solution and after each addition of HCl

a) NaNO2 dissociates into ions: Na+ + NO2- and reacts with water:

• NO2- is conj. base of weak acid HNO2

Step 1: ICE Chart The Kb expression for NO2-  is:

$\overline{){{\mathbf{K}}}_{{\mathbf{b}}}{\mathbf{=}}\frac{\mathbf{products}}{\mathbf{reactants}}{\mathbf{=}}\frac{\mathbf{\left[}{\mathbf{HNO}}_{\mathbf{2}}\mathbf{\right]}\mathbf{\left[}{\mathbf{OH}}^{\mathbf{-}}\mathbf{\right]}}{\mathbf{\left[}{\mathbf{NO}}_{\mathbf{2}}\mathbf{-}\mathbf{\right]}}}$

Calculate Kb:

Assuming the reaction is at 25°C:  Kw = 1.0x10-14

$\frac{{\mathbf{K}}_{\mathbf{w}}}{{\mathbf{K}}_{\mathbf{a}}}\mathbf{=}\frac{\overline{){\mathbf{K}}_{\mathbf{a}}}\mathbf{·}{\mathbf{K}}_{\mathbf{b}}}{\overline{){\mathbf{K}}_{\mathbf{a}}}}\phantom{\rule{0ex}{0ex}}\overline{){{\mathbf{K}}}_{{\mathbf{b}}}{\mathbf{=}}\frac{{\mathbf{K}}_{\mathbf{w}}}{{\mathbf{K}}_{\mathbf{a}}}}\phantom{\rule{0ex}{0ex}}{\mathbf{K}}_{\mathbf{b}}\mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{14}}}{\mathbf{4}\mathbf{.}\mathbf{6}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}}$

Kb = 2.17 x 10-11

Step 2: Solve for x = [|OH-]

${\mathbf{K}}_{{\mathbf{b}}}\mathbf{=}\frac{\mathbf{\left[}{\mathbf{HNO}}_{\mathbf{2}}\mathbf{\right]}\mathbf{\left[}{\mathbf{OH}}^{\mathbf{-}}\mathbf{\right]}}{\mathbf{\left[}{\mathbf{NO}}_{\mathbf{2}}\mathbf{-}\mathbf{\right]}}$

we can ignore x in the denominator

x = [OH-]  = 1.47 x 10-6 M

Step 3: Calculate pOH:

pOH = 5.83

89% (67 ratings) ###### Problem Details

Knowing that NaNO2 is the salt of a strong base and weak acid calculate :

a: pH of 0.100 M NaNO2 solution. Ka of HNO2 = 4.6 x 10-4.

b: pH of the solution obtained on adding 20.0 ml of a 0.100 M HCl solution to 60.0 ml of a 0.100 M NaNO2 .

c: pH of the solution obtained on adding 120.0 ml of a 0.100 M HCl solution to 60.0 ml of a 0.100 M NaNO2