21.7mL of 1.50M H2SO4 contains .0217*1.50 = .03255 moles of H2SO4
Each mole of H2SO4 reacts with 2 moles of KOH.
So, the 40mL of KOH contained 0.0651 moles.
.0651mole/.040L = 1.6275M
2KOH(aq)+H2SO4(aq)βK2SO4(aq)+2H2O(l)
Each mole of H2SO4 reacts with 2 moles of KOH.
So, the 40mL of KOH contained 0.0651 moles.
.0651mole/.040L = 1.6275M
So, let's break it down. We have 21.7 mL of 1.50 M H2SO4 that reacted with 40.0 mL of KOH. According to the balanced equation, itβs a 2:1 ratio between H2SO4 and KOH.
Now, let me put on my funny math hat. We can use the equation:
(Molarity of H2SO4) Γ (Volume of H2SO4) = (Molarity of KOH) Γ (Volume of KOH)
Plugging in the numbers, we get:
(1.50 M) Γ (21.7 mL) = (Molarity of KOH) Γ (40.0 mL)
Doing some quick math juggling, we find that the Molarity of KOH is:
Molarity of KOH = (1.50 M Γ 21.7 mL) / 40.0 mL
Calculating that, we find the molarity of KOH to be... drumroll, please...
0.815625 M
So there you have it! The molarity of the KOH solution is approximately 0.815625 M. Tada!
The balanced equation for the reaction is:
2KOH(aq) + H2SO4(aq) β K2SO4(aq) + 2H2O(l)
From the given equation, we can see that the mole ratio between KOH and H2SO4 is 2:1.
First, we need to determine the number of moles of H2SO4 used in the titration. To do this, we'll use the formula:
moles = molarity Γ volume
Given:
Volume of H2SO4 (V1) = 21.7 mL = 0.0217 L
Molarity of H2SO4 (M1) = 1.50 M
Using the formula, we can calculate the number of moles of H2SO4 as follows:
moles of H2SO4 = M1 Γ V1
= 1.50 M Γ 0.0217 L
= 0.03255 moles
Since the stoichiometric ratio between KOH and H2SO4 is 2:1, we can conclude that the number of moles of KOH is also 0.03255 moles.
Next, we'll find the volume of the KOH solution (V2) in liters, using the formula:
V2 = moles / molarity
Given:
Moles of KOH (n) = 0.03255 moles
Volume of KOH solution (V2) = 40.0 mL = 0.0400 L (converted to liters)
Using the formula, we can calculate the molarity (M2) of the KOH solution as follows:
Molarity of KOH solution (M2) = n / V2
= 0.03255 moles / 0.0400 L
= 0.81375 M
Therefore, the molarity of the aqueous KOH solution is approximately 0.81375 M.
1. Write down the balanced chemical equation:
2KOH(aq) + H2SO4(aq) β K2SO4(aq) + 2H2O(l)
2. Determine the stoichiometric ratio:
From the balanced equation, we can see that the stoichiometric ratio between KOH and H2SO4 is 2:1. This means that 2 moles of KOH react with 1 mole of H2SO4.
3. Convert the given volume and molarity of H2SO4 to moles:
The given volume of H2SO4 is 21.7 mL, which can be converted to liters by dividing by 1000: 21.7 mL Γ· 1000 mL/L = 0.0217 L.
The molarity of H2SO4 is given as 1.50 M.
Using the formula Molarity (M) = Moles (mol) / Volume (L), we can calculate the moles of H2SO4:
Moles of H2SO4 = Molarity Γ Volume = 1.50 M Γ 0.0217 L = 0.03255 mol
4. Use stoichiometry to find the moles of KOH:
Since the stoichiometric ratio between KOH and H2SO4 is 2:1, we can multiply the moles of H2SO4 by 2 to find the moles of KOH:
Moles of KOH = 2 Γ Moles of H2SO4 = 2 Γ 0.03255 mol = 0.0651 mol
5. Calculate the molarity of KOH:
Molarity of KOH = Moles of KOH / Volume of KOH solution (in liters)
We're given that the volume of KOH solution is 40.0 mL, which is 0.0400 L.
Molarity of KOH = 0.0651 mol / 0.0400 L = 1.62875 M
Therefore, the molarity of the KOH solution is approximately 1.63 M.