What volume (L) of 0.256 M potassium hydroxide solution would just neutralize 61.3 ml of 0.167 M H2SO4 solution?
To find the volume of potassium hydroxide (KOH) solution that would neutralize a given volume of sulfuric acid (H2SO4) solution, we can use the equation:
M1V1 = M2V2
Where:
M1 = molarity of acid (H2SO4) = 0.167 M
V1 = volume of acid (H2SO4) = 61.3 mL
M2 = molarity of base (KOH) = 0.256 M
V2 = volume of base (KOH) we need to find
First, we need to convert the volumes from milliliters (mL) to liters (L):
V1 = 61.3 mL = 61.3 mL / 1000 mL/L = 0.0613 L
Now, we can substitute the known values into the equation:
(0.167 M)(0.0613 L) = (0.256 M)(V2)
0.0102271 M⋅L = 0.256 M⋅V2
Next, we can solve for V2 by isolating it on one side of the equation:
V2 = 0.0102271 M⋅L / 0.256 M
V2 = 0.039957 L
Therefore, the volume of the 0.256 M potassium hydroxide solution required to neutralize 61.3 mL of the 0.167 M sulfuric acid solution is approximately 0.039957 liters, or 39.96 mL (rounded to two decimal places).