# How many grams of NaCl should be weighed to prepare 1 L of 20 ppm solution of Na^+?

Jul 10, 2015

You need 0.05 g of sodium chloride.

#### Explanation:

In order to determine how much sodium chloride must be dissolved in 1 L of water, start from the definition of parts per million, ppm.

A concentration of 1 ppm is equivalent to 1 part solute, in your case sodium chloride, for every 1 million parts solvent, in your case water.

To get a solution's concentration in ppm, you multiply the ratio that exists between the mass of the solute and the mass of the water by 1 million, or ${10}^{6}$.

This is the exact same approach you use when calculating percentage, the only difference being the fact that you need to multiply the ratio by 1 million, instead of by 100.

So, you can safely assume the density of water to be equal to $\text{1 g/mL}$. This would make the mass of water equal to

1cancel("L") * (1000cancel("mL"))/(1cancel("L")) * "1 g"/(1cancel("mL")) = "1000 g"

Sodium chloride will dissociate completely in aqueous solution to form sodium cations, $N {a}^{+}$, and chloride anions, $C {l}^{-}$.

$N a C {l}_{\left(a q\right)} \to N {a}_{\left(a q\right)}^{+} + C {l}_{\left(a q\right)}^{-}$

The important thing to notice here is that you have a $1 : 1$ mole ratio between sodium chloride and sodium cations. This mole ratio will help you determine the mass of sodium chloride you need to dissolve in order to get this particular solution.

The concentration of the target solution (in ppm) will be

$\text{ppm" = m_"solute"/m_"water} \cdot {10}^{6}$

Plug in your values and solve for ${m}_{\text{solute}}$.

m_"solute" = ("ppm" * m_"water")/10^6

${m}_{\text{solute" = (20 * "1000 g")/10^6 = "0.02 g}}$

This means that your solution must contain 0.02 g of sodium cations, $N {a}^{+}$. Use sodium's molar mass to determine how many moles of sodium cations would be present

0.02cancel("g") * "1 mole"/(23.0cancel("g")) = "0.000870 moles" $N {a}^{+}$

The aforementioned mole ratio tells you that you must add the same number of moles of sodium chloride to the solution.

0.000870cancel("moles"Na^(+)) * ("1 mole"NaCl)/(1cancel("mole"Na^(+))) = "0.000870 moles" $N a C l$

Now use sodium chloride's molar mass to determine how much you need

$0.000870 \cancel{\text{moles") * "58.44 g"/(1cancel("mole")) = color(green)("0.05 g NaCl}}$

So, if you dissolve 0.05 g of sodium chloride in 1 L of water you'll get a 20-ppm $N {a}^{+}$ solution.