A 0.2299 N acid is prepared by mixing 100 ml of 0.1234 N H2SO4 and 0.3200 N HCl. How much HCl is needed in the preparation of the solution?
118.2
Let x = amount of 0.3200 N HCl needed.
m.e. = # milliequivelents
Then m.e. H2SO4 + m.e. HCl = m.e. 0.2299 solution.
(100mL*0.1234N) + (XmL*0.3200N) = (100mL + xmL)*0.2299
Solve for X.
To find out how much HCl is needed in the preparation of the solution, we can use the concept of molarity and volume.
We have two solutions: 0.1234 N H2SO4 and 0.3200 N HCl.
First, let's calculate the moles of H2SO4 present in the solution.
Moles of H2SO4 = Molarity of H2SO4 * Volume of H2SO4 in liters
Given:
Molarity of H2SO4 = 0.1234 N
Volume of H2SO4 = 100 ml = 0.1 liters
Moles of H2SO4 = 0.1234 N * 0.1 L = 0.01234 moles
Similarly, let's calculate the moles of HCl present in the solution.
Moles of HCl = Molarity of HCl * Volume of HCl in liters
Given:
Molarity of HCl = 0.3200 N
Volume of HCl = unknown (let's call it "x" liters)
Moles of HCl = 0.3200 N * x L = 0.3200x moles
Now, since both H2SO4 and HCl react in a 1:1 ratio, we can equate the moles of H2SO4 to the moles of HCl.
0.01234 moles H2SO4 = 0.3200x moles HCl
Now, solve this equation to find the value of "x," which represents the volume of HCl in liters.
0.01234 / 0.3200 = x
x ≈ 0.0384375 liters
Since the volume was given in milliliters, we need to convert liters to milliliters.
x ≈ 0.0384375 * 1000 ml
x ≈ 38.4375 ml
Therefore, approximately 38.44 ml of HCl is needed in the preparation of the solution.