If a solution of Sulfuric acid (H2SO4) has a concentration of 2.1E-4M, calculate the concentration of its hydroxide ion. a. 4.2E-4M ...

Determine the concentration of `~H^+`

Sulfuric acid (`~H_2SO_4` ) is a strong acid and disassociates completely in water to form hydrogen and sulfate ions. The disassociation of sulfuric acid in water is shown below:

   `~H_2SO_4` -> `~2H^+` + `~SO_4^2^-`

Since 1 mole of `~H_2SO_4` disassociates into 2 moles of `~2H^+` , the concentration of hydrogen ions (`~H^+` ) is equal to twice the concentration of `~H_2SO_4` :

   2 x (2.1 x `~10^-^4` )...

Determine the concentration of `~H^+`

Sulfuric acid (`~H_2SO_4` ) is a strong acid and disassociates completely in water to form hydrogen and sulfate ions. The disassociation of sulfuric acid in water is shown below:

   `~H_2SO_4` -> `~2H^+` + `~SO_4^2^-`

Since 1 mole of `~H_2SO_4` disassociates into 2 moles of `~2H^+` , the concentration of hydrogen ions (`~H^+` ) is equal to twice the concentration of `~H_2SO_4` :

   2 x (2.1 x `~10^-^4` ) = 0.00042M `~H^+`

Choose a formula

We can use the following formula to calculate the [`~OH^-` ]:

   [`~H^+` ][`~OH^-` ] = 1.0 x `~10^-^14` M

1.0 x `~10^-^14`is the ionization constant for water (`~K_w` )

Rearrange the formula to isolate [`~OH^-` ] by dividing both sides by [`~H^+` ]:

   [`~OH^-` ] = 1.0 x `~10^-^14` /[`~H^+` ]

Solve for [`~OH^-` ] using [`~H^+` ]

  [`~OH^-` ] = 1.0 x `~10^-^14` /0.00042

               = 2.4 x `~10^-^11` M``

Therefore, the answer is c.