# Calculating the molarity of an acid solution

Concentrated hydrochloric acid is $37\% \ce{HCl}$ by mass and has a density of $1.2\rm~\frac{g}{ml}$. Calculate the molarity of a solution made by diluting $125\rm~ ml$ of concentrated $\ce{HCl}$ with water to a total volume of $2\rm~ L$.

I got $194.88\rm~M$ but that's not correct. First, I found the molarity of the $37\%$ solution to be $12.18\rm~M$; then I used the $M_1V_1=M_2V_2$ formula.

Finding Molarity $$\rm\frac{100g}{1.20~\frac{g}{mL}}= 83.3~mL = 0.083~L$$ $$\rm\frac{37~g~HCl}{36.46~\frac{g}{mol}} = 1.015~mol~HCl$$

$$\rm\frac{1.015~mol}{0.083~L}= 12.18~\frac{mol}{L}$$ (This answer was marked correct on my hw)

Finding molarity of concentrated HCl $$M_1V_1=M_2V_2$$ $$M_1\times(0.125)=(12.18)(2)$$ $$M_1=194.88~\rm\frac{mol}{L}$$

You had the correct formula but plugged in the numbers wrong. For dilutions the formula is: $$\mathrm{M}_1 \mathrm{V}_1 = \mathrm{M}_2 \mathrm{V}_2$$ where
• $\mathrm{M}_1$ is the original molar strength
• $\mathrm{V}_1$ is the original volume
• $\mathrm{M}_2$ is the diluted molar strength
• $\mathrm{V}_1$ is the diluted volume
The formula works because $$\mathrm{M} \mathrm{V} = \mathrm{(M)(L) = (moles/L)(L) = moles}$$ so regardless of how much dilution you have the same number of moles of reagent.
So for this particular problem the final part of the solution is: $$\mathrm{(12.18M)(0.125L)}=(x)\mathrm{(2L)}$$ or $$x = \dfrac{\mathrm{(12.18M)(0.125L)}}{\mathrm{(2L)}} = \mathrm{0.761 M}$$