How do we make a #1*L# volume of a solution that is #0.5*mol*L^-1# with respect to #HCl# from a solution that is #10.6*mol*L^-1# with respect to #HCl(aq)#?

Question

Cameron Andrews · Accepted Answer

Given proper attire, with conc. acid THE PRIMARY PRACTICAL CONSIDERATION IS #"ADD YOUR ACID TO WATER"# and NEVER #"VICE VERSA"#. We need approx. a #50*mL# volume of conc. acid.

And why so? #"Because if you spit in acid, it spits back!"# I kid you not...... See this older answer. And other practical considerations. Wear a pair of SAFETY SPECTACLES to protect your mince pies; and wear a lab coat to protect your clothing......You should do this automatically in a lab (if you are a speccy, your prescription glasses are an adequate protection). Most chemists upon entering a lab will IMMEDIATELY put on a pair of safety spex from a tray, or put on the safety spec that they carry in their pocket. This is certainly a habit to develop. And to make a #1*L# volume of #0.5*mol*L^-1# #HCl#; we use the relationship..... #C_1V_1=C_2V_2#, and we solve for #V_1#..... #V_1=(C_2V_2)/C_1=(0.5*mol*L^-1xx1*L)/(10.6*mol*L^-1)~=50*mL#. And remember to add the #"WATER to YOUR ACID..............."#

Cooper Anderson · Answer

undefined To make a 1 L volume of a solution that is 0.5 mol/L with respect to HCl from a solution that is 10.6 mol/L with respect to HCl(aq), you would dilute the concentrated solution by adding water. The dilution formula is: M1V1 = M2V2, where M1 is the initial concentration, V1 is the initial volume, M2 is the final concentration, and V2 is the final volume. Rearranging the formula, you get: V1 = (M2 * V2) / M1. Plugging in the values, you get: V1 = (0.5 mol/L * 1 L) / 10.6 mol/L = 0.047 L. So, you would need to measure 47 mL of the concentrated solution and add enough water to bring the total volume to 1 L to obtain a solution with a concentration of 0.5 mol/L HCl.