Question

Picric acid has been used in the leather industry and in etching copper. However, its laboratory use has been restricted because it dehydrates on standing and can become shock sensitive. It has an acid dissociation constant of 0.42.
a. What is the [H3O+] for an 0.52 M solution of picric acid? Enter to 4 decimal places.

Answer 1

Answer:

0.3023 M

Explanation:

Let Picric acid = $H_{picric}$

So,  $H_{picric}$     +       $H_2}O$          ⇄      $H_3}O^+$     +     $Picric^-$

The ICE table can be given as:

                          $H_{picric}$     +       $H_2}O$          ⇄      $H_3}O^+$     +     $Picric^-$

Initial:                0.52                                               0                  0

Change:             - x                                                 + x                 + x

Equilibrium:      0.52 - x                                        + x                 + x

Given that;

acid dissociation constant  ($K_a$) = 0.42

$K_a = \frac{[H_3O^+][Picric^-]}{H_{picric}}$

$0.42 = \frac{[x][x]}{0.52-x}}$

$0.42 = \frac{[x]^2}{0.52-x}}$

0.42(0.52-x) = x²

0.2184 - 0.42x = x²

x²  + 0.42x - 0.2184 = 0                   -------------------- (quadratic equation)

Using the quadratic formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$    ;     ( where +/-  represent ± )

= $\frac{-0.42+/-\sqrt{(0.42)^2-4(1)(-0.2184)} }{2*1}$

= $\frac{-0.42+/-\sqrt {0.1764+0.8736} }{2}$

= $\frac{-0.42+\sqrt {1.0496} }{2}$     OR   $\frac{-0.42-\sqrt {1.0496} }{2}$

= $\frac{-0.42+1.0245}{2}$       OR    $\frac{-0.42-1.0245}{2}$

= $\frac{0.6045}{2}$                 OR    $-\frac{1.4445}{2}$

= 0.30225          OR     - 0.72225

So, we go by the +ve integer that says:

x =  0.30225

x = [ $H_3}O^+$ ] = [   $Picric^-$ ] =  0.3023  M

∴  the value of  [H3O+] for an 0.52 M solution of picric acid  = 0.3023 M     (to 4 decimal places).