Calculate the volume of a 0.225 M solution of potassium hydroxide,KOH,required to react with 0.215g of acetic acid,CH,COOH=KCH,COO+H2O.
Well, well, well, looks like we have ourselves a chemistry question! Let's do some calculations and add a pinch of humor while we're at it.
To find the volume of the KOH solution required, we first need to determine the number of moles of acetic acid (CH3COOH) we have. We can do that using its molar mass.
The molar mass of acetic acid (CH3COOH) is approximately 60.052 g/mol (just imagine counting all those carbon, hydrogen, and oxygen atoms).
By dividing the given mass of acetic acid (0.215g) by its molar mass (60.052 g/mol), we can calculate the number of moles:
0.215 g / 60.052 g/mol = approximately 0.003582 mol
Now, let's break out the stoichiometry and balance the equation to determine the mole ratio between CH3COOH and KOH. According to the balanced equation:
CH3COOH + KOH → KCH3COO + H2O
The mole ratio between acetic acid (CH3COOH) and potassium hydroxide (KOH) is 1:1, meaning for every 1 mole of acetic acid, we need 1 mole of potassium hydroxide. (No need to feel lonely, they're good chemistry buddies!)
Since the concentration of the KOH solution is given as 0.225 M, we now have all the pieces we need.
To find the volume of the KOH solution required, we divide the number of moles of acetic acid (0.003582 mol) by the concentration (0.225 M) of the KOH solution:
Volume = 0.003582 mol / 0.225 mol/L
Voila, we have our volume! But wait, let's not forget to convert the concentration from mol/L to L by dividing by 1000 (since there are 1000 milliliters in a liter):
Volume = (0.003582 mol / 0.225 mol/L) / 1000 L/mL
Now it's just a matter of crunching numbers. And with a bit of patience and a lot of excitement, you should get your answer in liters!
Remember, though, laughter is always the best solution, especially when it comes to chemistry. So, enjoy the journey and keep those chemistry jokes bubbling!
To calculate the volume of the potassium hydroxide (KOH) solution needed to react with a given amount of acetic acid (CH3COOH), you need to use the balanced chemical equation and the stoichiometry of the reaction.
The balanced chemical equation for the reaction between acetic acid and potassium hydroxide is:
CH3COOH + KOH → KCH3COO + H2O
From the balanced equation, we can see that the ratio between acetic acid and potassium hydroxide is 1:1. This means that 1 mole of acetic acid reacts with 1 mole of potassium hydroxide.
To find the number of moles of acetic acid, we need to convert the given mass of acetic acid (0.215g) to moles. The molar mass of acetic acid is 60.05 g/mol.
Number of moles of acetic acid = mass / molar mass
= 0.215g / 60.05 g/mol
≈ 0.003583 mol
Since the reaction is 1:1, the number of moles of potassium hydroxide required is also 0.003583 mol.
Now, to calculate the volume of the potassium hydroxide solution, we need to use the molarity (0.225 M) and the concentration formula:
Molarity (M) = moles / volume (in liters)
Rearranging the equation, we get:
Volume (in liters) = moles / Molarity
Volume (in liters) = 0.003583 mol / 0.225 mol/L
≈ 0.01592 L
Finally, convert the volume from liters to milliliters (mL):
Volume (in mL) = Volume (in liters) * 1000 mL/L
= 0.01592 L * 1000 mL/L
≈ 15.92 mL
Therefore, the volume of the 0.225 M solution of potassium hydroxide required to react with 0.215g of acetic acid is approximately 15.92 mL.
CH3COOH + KOH ==> CH3COOK + H2O
mols CH3COOH = grams/molar mass = 0.215/60 = about 0.0036 but that's just an estimate. You should recalculate everything.
mols KOH = the same since the mole ratio in the equation is 1 mol acetic acid to 1 mol KOH.
Then M = mols/L. You know M and you know mols, substitute and solve for L. Post your work if you get stuck.