How much 0.200 M KOH is required to completely neutralize 25.0 mL of 0.150 M HClO4?
molesKOH=molesHClO4
volume*0.2=.025*.150
solve for volume in liters
18.75L
Why did the acid go to therapy? Because it had some unresolved base issues! Now, let's calculate the amount of KOH needed to neutralize the HClO4.
To do this, we can use the equation:
M1V1 = M2V2
Where:
M1 = the concentration of KOH
V1 = the volume of KOH
M2 = the concentration of HClO4
V2 = the volume of HClO4
So, by plugging in the values we have:
(0.200 M)(V1) = (0.150 M)(25.0 mL)
Let's solve it together:
0.200V1 = 0.150(25.0)
0.200V1 = 3.75
V1 = 3.75 / 0.200
V1 = 18.75 mL
Therefore, approximately 18.75 mL of 0.200 M KOH is required to completely neutralize 25.0 mL of 0.150 M HClO4.
To determine the amount of 0.200 M KOH required to completely neutralize 25.0 mL of 0.150 M HClO4, you can use the equation:
M1V1 = M2V2
where:
M1 = molarity of the first solution (KOH)
V1 = volume of the first solution (KOH) in liters
M2 = molarity of the second solution (HClO4)
V2 = volume of the second solution (HClO4) in liters
First, convert the volume of HClO4 from milliliters to liters:
V2 = 25.0 mL / 1000 = 0.025 L
Next, substitute the known values into the equation. Let's assume x is the volume of KOH required in liters:
(0.200 M)(x) = (0.150 M)(0.025 L)
Simplify the equation:
0.200x = 0.150 * 0.025
0.200x = 0.00375
Now, solve for x:
x = 0.00375 / 0.200
x ≈ 0.01875 L
Finally, convert the volume of KOH from liters to milliliters:
0.01875 L * 1000 = 18.75 mL
Therefore, approximately 18.75 mL of 0.200 M KOH is required to completely neutralize 25.0 mL of 0.150 M HClO4.
To determine the amount of KOH required to neutralize the HClO4 solution, you need to use the equation:
M1V1 = M2V2
Where:
M1 = molarity of the first solution (HClO4)
V1 = volume of the first solution (HClO4)
M2 = molarity of the second solution (KOH)
V2 = volume of the second solution (KOH)
In this case:
M1 = 0.150 M (HClO4)
V1 = 25.0 mL (HClO4) = 0.025 L (HClO4)
M2 = 0.200 M (KOH)
V2 = unknown (KOH)
Substituting these values into the equation:
(0.150 M)(0.025 L) = (0.200 M)(V2)
0.00375 mol = 0.200 M * V2
V2 = 0.00375 mol / 0.200 M
V2 = 0.01875 L
Lastly, convert the volume from liters to milliliters:
V2 = 0.01875 L * 1000 mL/L
V2 = 18.75 mL
So, approximately 18.75 mL of 0.200 M KOH is required to completely neutralize 25.0 mL of 0.150 M HClO4.