g What is the molarity of hydrochloric acid if 40.95 mL of HCl is required to neutralize 0.550 g of sodium oxalate, Na2C2O4
1 answer:
Answer:
0.0002 M
Explanation:
<em>The molarity of the HCl required would be 0.0002 M.</em>
First, let us consider the balanced equation of the reaction:
<em>Stoichiometrically, 1 mole of </em><em> reacts with 2 moles of </em>
<em> for a complete neutralization reaction.</em>
Recall that: mole =
Mole of 0.550 g sodium oxalate = 0.550/134 = 0.0041 mole
<em>If 1 mole </em><em> requires 2 moles HCl, then 0.0041 mole will require</em>:
0.0041 x 2 = 0.0082 mole HCl
Volume of the HCl = 40.95 L
Molarity = mole/volume
Hence, molarity of the HCl = 0.0082/40.95 = 0.0002 M
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This is a straightforward dilution calculation that can be done using the equation
where <em>M</em>₁ and <em>M</em>₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and <em>V</em>₁ and <em>V</em>₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.
Here, we have the initial concentration (<em>M</em>₁) and the initial (<em>V</em>₁) and final (<em>V</em>₂) volumes, and we want to find the final concentration (<em>M</em>₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for <em>M</em>₂:
Substituting in our values, we get
So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.