Titration of 25.00 mL of KOH required 200.00 mL of 0.0050 M acetic acid, CH3COOH. What is the molarity (M) of the KOH solution? Please show calcutions.
HAc = CH3COOH = acetic acid
KOH = potassium hydroxide
mL KOH x M KOH = mL HAc x M HAc
Substitute and solve for M KOH
0.04
To determine the molarity (M) of the KOH solution, we can use the equation:
M1V1 = M2V2
where:
M1 = Molarity of acetic acid
V1 = Volume of acetic acid
M2 = Molarity of KOH
V2 = Volume of KOH
In this case, we have:
M1 = 0.0050 M (given)
V1 = 200.00 mL (given)
V2 = 25.00 mL (given)
Now we can plug those values into the equation and solve for M2:
M1V1 = M2V2
(0.0050 M)(200.00 mL) = M2(25.00 mL)
To simplify, let's convert the volumes from milliliters to liters:
(0.0050 M)(0.200 L) = M2(0.0250 L)
0.0010 moles = M2(0.0250 L)
Now, solve for M2 (molarity of KOH):
M2 = 0.0010 moles / 0.0250 L
M2 = 0.040 M
Therefore, the molarity (M) of the KOH solution is 0.040 M.