How can I dilute 13% Hydrochloric Acid to 5%? [closed]








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This is likely too simple of a question, but I just want to make sure I'm understanding things correctly.


I have 1 liter of 13% Hydrochloric Acid solution. I need 5% solution. Is it correct that I can just add the proportional amount of distilled (?) water to end up with a 5% solution? (add 1.6 liters of distilled water?)




In order not to waste your resources I recommend that you specify the quantity that you need for the upcoming reaction.


To make a diluted quantity of $x$ liters you may use the equation $$c\_1\*v\_1 = c\_2\*v\_2 \\ v\_2 = \frac{c\_1\*v\_1}{c\_2} $$


Where $v\_2$ is the volume that you want to use for the reaction, $v\_1$ is the volume from the concentrated solution that you need to take, $c\_1$ is the concentration of the concentrated solution, and $c\_2$ is the concentration of the solution that you want to use.


For your solution you may use the weight percentages or convert them to molarities if you want to use volumetric lab glassware.


To convert the weight percentages to molarities, use the equation $$c = \frac{x\*\rho}{m\*(100-x)}$$


Where x is the weight percentage of the solution so for 5% you use the number 5 in the equation, m is the molar mass of the hydrochloric acid. To facilitate calculating the molar masses of different compounds I use the on-line tool [Molar Mass Calculator](http://www.webqc.org/mmcalc.php).


To get the best results I recommend using a volumetric flask and substituting the volume for the volumetric flask instead of the volume of the acid that you want to use. So the equation that want to use is $$v\_{needed} = \frac{c\_{5\%}\*v\_{volumetric}}{c\_{13\%}} $$
You then put the volume that you obtained in the volumetric flask and fill the rest with distilled water till the mark and you have thus obtained your solution.


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