What volume of .01M KOH solution is needed to react with 20ml .01M Zn(OH)2 solution
To find out the volume of 0.01 M KOH solution required to react with 20 mL of 0.01 M Zn(OH)2 solution, we need to determine the stoichiometry of the reaction between these two compounds.
The balanced chemical equation for the reaction between KOH and Zn(OH)2 is as follows:
2 KOH + Zn(OH)2 -> Zn(OH)4 + 2 KOH
From this equation, we can infer that 2 moles of KOH react with 1 mole of Zn(OH)2. Therefore, the molar ratio between KOH and Zn(OH)2 is 2:1.
First, let's calculate the number of moles of Zn(OH)2 in 20 mL of the 0.01 M solution:
Moles of Zn(OH)2 = Concentration (M) × Volume (L)
= 0.01 M × 0.020 L
= 0.0002 moles
Since the molar ratio between KOH and Zn(OH)2 is 2:1, we need twice the number of moles of Zn(OH)2 to react with KOH. Thus, let's calculate the number of moles of KOH required:
Moles of KOH = 2 × Moles of Zn(OH)2
= 2 × 0.0002 moles
= 0.0004 moles
Now that we have the number of moles required, let's calculate the volume of 0.01 M KOH solution needed:
Volume (L) = Moles / Concentration (M)
= 0.0004 moles / 0.01 M
= 0.04 L
To convert the volume from liters to milliliters:
Volume (mL) = Volume (L) × 1000
= 0.04 L × 1000
= 40 mL
Therefore, you need 40 mL of 0.01 M KOH solution to react with 20 mL of 0.01 M Zn(OH)2 solution.