A solution is prepared from 8 grams of acetic acid (CH2CO2H) and 725 grams of water. What is the molarity of this solution?
2 answers:
Hello!
First, we can assume the density of water to be 1 g/mL, so the volume of water would be 725 mL. The density of Acetic Acid (Pure) is 1,05 g/mL so 8 grams would represent 7,61 mL
Now we can apply the following conversion factor to calculate the molarity of the solution, using the molar mass of Acetic Acid:
![[CH_3CO_2H]=\frac{8 g_{CH_3CO_2H} }{725 mL + 7,61 mL}* \frac{1000 mL}{1 L}* \frac{1 mol_{CH_3CO_2H} }{60,05 g_{CH_3CO_2H} }](https://tex.z-dn.net/?f=%5BCH_3CO_2H%5D%3D%5Cfrac%7B8%20g_%7BCH_3CO_2H%7D%20%20%7D%7B725%20mL%20%2B%207%2C61%20mL%7D%2A%20%5Cfrac%7B1000%20mL%7D%7B1%20L%7D%2A%20%5Cfrac%7B1%20mol_%7BCH_3CO_2H%7D%20%7D%7B60%2C05%20g_%7BCH_3CO_2H%7D%20%7D)
![[CH_3CO_2H]=0,1818 M](https://tex.z-dn.net/?f=%5BCH_3CO_2H%5D%3D0%2C1818%20M)
So, the concentration of acetic acid would be 0,1818 M
Have a nice day!
First, we can assume the density of water to be 1 g/mL, so the volume of water would be 725 mL. The density of Acetic Acid (Pure) is 1,05 g/mL so 8 grams would represent 7,61 mL
Now we can apply the following conversion factor to calculate the molarity of the solution, using the molar mass of Acetic Acid:
So, the concentration of acetic acid would be 0,1818 M
Have a nice day!
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Answer:
0.3023 M
Explanation:
Let Picric acid =
So, +
⇄
+
The ICE table can be given as:
+
⇄
+
Initial: 0.52 0 0
Change: - x + x + x
Equilibrium: 0.52 - x + x + x
Given that;
acid dissociation constant () = 0.42
0.42(0.52-x) = x²
0.2184 - 0.42x = x²
x² + 0.42x - 0.2184 = 0 -------------------- (quadratic equation)
Using the quadratic formula;
; ( where +/- represent ± )
=
=
= OR
= OR
= OR
= 0.30225 OR - 0.72225
So, we go by the +ve integer that says:
x = 0.30225
x = [ ] = [
] = 0.3023 M
∴ the value of [H3O+] for an 0.52 M solution of picric acid = 0.3023 M (to 4 decimal places).