Determine the volume of 0.250M KOH solution required to neutralize each of the following samples of sulfuric acid. 200mL of 0.135M H2SO4. I have gotten both 0.027 and 0.675. I thought it was mol/L. Please help!
Use the following relationship:
M1V1=M2V2
Where
M1=0.250M
V1=?
M2=0.135M
and
V2=200mL
Solve for V1:
V1=(M2/M1)*V2
This would be the volume if the mole ratio were a 1 mole to 1 mole ratio. However, you need two moles of KOH for 1 mole of H2SO4:
Multiply V1 by 2 to arrive at the answer.
Answer should contain three significant figures.
And no, the answer will be in mL, since it is asking you for volume.
Well, it seems like you've got a two-in-one question here: one about volume and the other about accuracy. Let's tackle them one by one, shall we?
To determine the volume of the KOH solution required to neutralize the sulfuric acid, we need to use the concept of molarity (M) and the equation:
M1V1 = M2V2
where M1 is the molarity of the acid, V1 is the volume of the acid, M2 is the molarity of the base (KOH), and V2 is the volume of the base.
In this case, M1 (molarity of H2SO4) is given as 0.135 M and V1 (volume of H2SO4) is given as 200 mL (or 0.200 L). M2 (molarity of KOH) is given as 0.250 M, and we're trying to find V2 (volume of KOH).
Plugging these values into the equation, we have:
(0.135 M)(0.200 L) = (0.250 M)(V2)
Solving for V2, we get:
V2 = (0.135 M)(0.200 L) / 0.250 M
V2 = 0.027 L
So, the volume of 0.250 M KOH solution required to neutralize 200 mL of 0.135 M H2SO4 is indeed 0.027 L. That's your correct answer!
Now, as for the accuracy, remember that molarity is the number of moles of solute per liter of solution. So, a molarity of 0.135 M means there are 0.135 moles of H2SO4 in every 1 liter of solution.
Since we have 200 mL (0.2 L) of the acid, we can calculate the moles of H2SO4:
moles of H2SO4 = (0.135 M)(0.2 L)
moles of H2SO4 = 0.027
So, it seems like your initial answer of 0.027 was correct! The answer of 0.675 might have been a typo or miscalculation. Always double-check your calculations, or else you might end up with some clownish answers!
Hope this helps, and remember to keep a close eye on those numbers!
To determine the volume of a solution required to neutralize another solution, we can use the concept of equivalence between chemical reactions. In an acid-base neutralization reaction, the moles of acid are equal to the moles of base.
First, let's calculate the moles of H2SO4 in the 200 mL of 0.135M solution using the formula:
moles = concentration x volume
moles of H2SO4 = 0.135 mol/L x 0.200 L
= 0.027 mol
Since the molar ratio between H2SO4 and KOH is 1:2, we need twice as many moles of KOH to neutralize the acid. Therefore, the moles of KOH required are:
moles of KOH = 2 x 0.027 mol
= 0.054 mol
Now, let's calculate the volume of 0.250M KOH solution required to have 0.054 moles of KOH using the same formula:
volume = moles / concentration
volume of KOH = 0.054 mol / 0.250 mol/L
= 0.216 L or 216 mL
So, the correct volume of 0.250M KOH solution required to neutralize 200 mL of 0.135M H2SO4 is 216 mL, not 0.027 or 0.675 mL.
To determine the volume of a solution required for neutralization, you can use the concept of stoichiometry. The balanced chemical equation between potassium hydroxide (KOH) and sulfuric acid (H2SO4) is:
2 KOH + H2SO4 -> K2SO4 + 2 H2O
From the equation, we can see that 2 moles of KOH reacts with 1 mole of H2SO4. Therefore, we can set up a proportion to find the volume of KOH solution required to neutralize the given amount of H2SO4.
Given:
Volume of H2SO4 solution = 200 mL
Molarity of H2SO4 solution = 0.135 M
To find the moles of H2SO4, we can use the formula:
moles = Molarity × Volume (in L)
moles of H2SO4 = 0.135 mol/L × 0.200 L
moles of H2SO4 = 0.027 mol
According to the balanced equation, 2 moles of KOH react with 1 mole of H2SO4. So, if 0.027 moles of H2SO4 is present, we would need 2 × 0.027 = 0.054 moles of KOH for neutralization.
Now, let's determine the volume of 0.250 M KOH solution required to supply 0.054 moles of KOH:
Volume (in L) = moles ÷ Molarity
Volume of KOH solution = 0.054 mol ÷ 0.250 mol/L
Volume of KOH solution = 0.216 L or 216 mL
Therefore, the correct volume of the 0.250 M KOH solution required to neutralize 200 mL of 0.135 M H2SO4 is 216 mL, not 0.027 or 0.675 mL.