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

Mar 4, 2015

0.009 mM.

Explanation:

The concentration of iodide, or ${\text{I}}^{-}$, will increase by $\text{0.009 mM}$.

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

A $\text{w/v}$ 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

$\text{12 mL" * "10 g"/"100 mL" = "1.2 g KI}$

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

$\text{1.2 g KI" * "1 mole"/"166.0 g" = "0.00723 moles KI}$

Now, because potassium iodide fully dissociates in aqueous solution into ${\text{K}}^{+}$ and ${\text{I}}^{-}$ ions, the number of moles of iodide will be

$K {I}_{\left(a q\right)} r i g h t \le f t h a r p \infty n s {K}_{\left(a q\right)}^{+} + {I}_{\left(a q\right)}^{-}$

${\text{0.00723 moles KI" * "1 mole I"^(-)/"1 mole KI" = "0.00723 moles I}}^{-}$

It is safe to assume that the added $\text{KI}$ 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

$C = \frac{n}{V} = \text{0.00723 moles I"^(-)/"800 L" = "0.00904 mM}$

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.