Question

How much would a dosage of 12ml 10% Potassium Iodide (KI) w/v solution increase Iodide levels in a total voluve of 800 liters?

Answer 1

0.009 mM.

Explanation:

The concentration of iodide, or `"I"^(-)`, will increase by `"0.009 mM"`.

First, start by figuring out exactly how much iodide you'll be adding to the tank. Use the given `"w/v "` percent concentration to determine how much potassium iodide you'd get in 12 mL.

A `"w/v"` percent contration solution is defined as

`"%w/v" = m_("solute")/V_("solution") * 100`

In your case, a 10% w/v solution would have 10 g of potassium iodide for every 100 mL of solution. As a result, the 12-mL sample will contain

`"12 mL" * "10 g"/"100 mL" = "1.2 g KI"`

Use potassium iodide's molar mass to determine how many moles you get from that mass

`"1.2 g KI" * "1 mole"/"166.0 g" = "0.00723 moles KI"`

Now, because potassium iodide fully dissociates in aqueous solution into `"K"^(+)` and `"I"^(-)` ions, the number of moles of iodide will be

`KI_((aq)) rightleftharpoons K_((aq))^(+) + I_((aq))^(-)`

`"0.00723 moles KI" * "1 mole I"^(-)/"1 mole KI" = "0.00723 moles I"^(-)`

It is safe to assume that the added `"KI"` solution will not have an impact on the volume of the tank

`V_("tank total") = "800 L" + 12 * 10^(-3)"L" = "800.012 L" ~= "800 L"`

Therefore, the added concentration of iodide in the tank will be

`C = n/V = "0.00723 moles I"^(-)/"800 L" = "0.00904 mM"`

Rounded to one sig fig, the number of sig figs in 800 L, the answer will be

`["I"^(-)] = "0.009 mM"`

SIDE NOTE. Keep in mind that the concentration is milimolar, not molar.