# Can you help me with this? Commercially available HCl contains 38% w/w (weight by weight) of acid. If density is 1.19 g/ml, calculate molality, molarity and normality of the acid solution.

Jul 8, 2015

Here's how you can approach such problems.

#### Explanation:

You know that your solution is 38% w/w hydrochloric acid. This means that every 100 g of solution will contain 38 g of acid.

To make the calculations easier, assume that you're dealing with a 1.00-L sample of stock solution.

Use the solution's density to determine what the mass of this sample would be

1.00cancel("L") * (1000cancel("mL"))/(1cancel("L")) * "1.19 g"/(1cancel("mL")) = "1190 g"

Now use the known percent concentration by mass to determine how many grams of hydrochloric acid you'd get

1190cancel("g solution") * "38 g HCl"/(100cancel("g solution")) = "452.2 g HCl"

To determine the solution's molarity, use hydrochloric acid's molar mass - this will get you the number of moles of acid present in the sample.

452.2cancel("g") * "1 mole"/(36.46cancel("g")) = "12.40 moles"

Since the sample has a volume of 1.00-L, the molarity will be

C = n/V = "12.40 moles"/"1.00 L" = color(green)("12.4 M")

To get the solution's molality, you need to know the mass of water. Since you know the mass of the solution and that of acid, you can write

${m}_{\text{sol" = m_"water" + m_"acid}}$

${m}_{\text{water" = m_"sol" - m_"acid" = 1190 - 452.2 = "737.8 g}}$

The molality will thus be - do not forget to convert the mass of water to kilograms!

b = n/m_"water" = "12.40 moles"/(737.8 * 10^(-3)"kg") = color(green)("16.8 molal")

Since hydrochloric acid, $H C l$, can only release one mole of protons per mole of acid in aqueous solution, the solution's normality will be equal to its molarity.

$\text{normality" = C/f_"eq}$, where

$C$ - the solution's molarity;
${f}_{\text{eq}}$ - the equivalence factor, in your case equal to 1.

"normality" = "12.40 M"/1 = color(green)("12.4 N")

SIDE NOTE I left the values rounded to three sig figs.