Chemistry Problem-Solving: Concentrated HCl Solutions
How can we determine the molarity of concentrated HCl solution?
A bottle of concentrated hydrochloric acid contains 36.0% HCl by mass and has a density of 1.18 g/mL. What is the molarity of the concentrated HCl?
What volume of the concentrated HCl solution is needed to prepare a 0.250 L of 2.00 M HCl solution through dilution?
How much of the concentrated HCl solution should be used to create the desired diluted solution?
If a spill of 1.75 L of concentrated HCl occurs, how much sodium hydrogen carbonate is required to neutralize the spill?
In the event of an acid spill, what mass of sodium hydrogen carbonate is necessary to safely neutralize the spillage?
Answers:
a) 11.64 M
b) 43 mL
c) 1.7 kg
The molarity of the concentrated HCl solution is approximately 11.64 M. To determine this, we first calculated the mass of HCl in 1L of solution by using the provided density and percentage composition.
We then found that to prepare 0.250 L of a 2.00 M HCl solution through dilution, approximately 43 mL of the concentrated HCl solution is required. This calculation was done using the formula C1V1 = C2V2 (M1V1 = M2V2).
In the case of a spill of 1.75 L of concentrated HCl, approximately 1.71 kg of sodium hydrogen carbonate would be needed to neutralize it. This calculation involved finding the moles of HCl in the spill and determining the molar mass of NaHCO₃ for the neutralization reaction.
Explanation:To find the molarity of the concentrated HCl solution, we first calculated the mass of HCl in 1L using the provided density and percentage composition. With this information, we determined that the molarity of the concentrated HCl solution is 11.64 M.
For the dilution process to prepare a 0.250 L of 2.00 M HCl solution, we used the formula M1V1 = M2V2 to find that approximately 43 mL of the concentrated HCl solution is needed.
In the event of a spill of 1.75 L of concentrated HCl, approximately 1.71 kg of sodium hydrogen carbonate would be required to neutralize the spill. This calculation considered the moles of HCl in the spill and the molar mass of NaHCO₃ for the neutralization reaction.