The volume of 0.025M Ca(OH)2 solution which can neutralise 100ml of0.0001M H3PO4
To determine the volume of the 0.025M Ca(OH)2 solution needed to neutralize 100ml of 0.0001M H3PO4, we can use the concept of stoichiometry.
First, let's write the balanced equation for the reaction between Ca(OH)2 and H3PO4:
Ca(OH)2 + 2H3PO4 → Ca(H2PO4)2 + 2H2O
From the balanced equation, we can see that 1 mole of Ca(OH)2 reacts with 2 moles of H3PO4.
Next, we can calculate the number of moles of H3PO4 present in 100ml of 0.0001M solution:
moles of H3PO4 = volume (L) × concentration (M)
= 0.1 L × 0.0001 M
= 0.00001 moles
Since 1 mole of Ca(OH)2 reacts with 2 moles of H3PO4, we know that 0.00001 moles of H3PO4 will react with half as many moles of Ca(OH)2, which is 0.000005 moles.
Now, we can calculate the volume of 0.025M Ca(OH)2 solution required:
moles of Ca(OH)2 = 0.000005 moles
volume (L) = moles / concentration
= 0.000005 moles / 0.025 M
= 0.0002 L = 0.2 ml
Therefore, to neutralize 100ml of 0.0001M H3PO4, you would need approximately 0.2 ml of the 0.025M Ca(OH)2 solution.
3Ca(OH)2 + 2H3PO4 ==> Ca3(PO4)2 + 6H2O
mols H3PO4 = M x L = ?
Using the coefficients in the balanced equation, convert mols H3PO4 to mols Ca(OH)2.
Now M Ca(OH)2 = mols Ca(OH)2/L Ca(OH)2. You know M and mols, solve for L.