Consider the following neutralisation reaction:
2 H3PO4(aq) + 3 Ca(OH)2 (aq) → Ca3(PO4)2 + 6 H2O
Determine the volume of 0.75 M H3PO4 necessary to neutralise 35 ml of 0.45 M Ca(OH)2
since 2 moles of H3PO4 reacts with 3 moles of Ca(OH)2, it takes 2/3 mole for each mole of Ca(OH)2
So, you have .035L*.45mole/L = .01575 moles of Ca(OH)2.
2/3 of that is 0.0105 moles of H3PO4.
.0105 mole / (.75 mole/L) = .014L = 14ml
To determine the volume of 0.75 M H3PO4 necessary to neutralize 35 mL of 0.45 M Ca(OH)2, we need to use the stoichiometry of the reaction.
Looking at the balanced equation:
2 H3PO4(aq) + 3 Ca(OH)2(aq) → Ca3(PO4)2 + 6 H2O
We can see that the stoichiometric ratio between H3PO4 and Ca(OH)2 is 2:3. This means that 2 moles of H3PO4 react with 3 moles of Ca(OH)2.
First, we need to determine the moles of Ca(OH)2 present in 35 mL of 0.45 M solution:
Moles of Ca(OH)2 = volume (in liters) × concentration
= 35 mL × 0.45 mol/L
= 0.01575 moles
According to the stoichiometry, the number of moles of H3PO4 required to neutralize the moles of Ca(OH)2 is in a 2:3 ratio.
Moles of H3PO4 = (2/3) × moles of Ca(OH)2
= (2/3) × 0.01575 moles
= 0.0105 moles
Finally, to calculate the volume of the 0.75 M H3PO4 solution required, we can use the equation:
Moles = volume (in liters) × concentration
0.0105 moles = volume (in liters) × 0.75 mol/L
Rearranging the equation, we can solve for the volume:
Volume (in liters) = moles / concentration
= 0.0105 moles / 0.75 mol/L
= 0.014 L
Since 1 L = 1000 mL, we can convert the volume to milliliters:
Volume (in mL) = 0.014 L × 1000 mL/L
= 14 mL
Therefore, the volume of the 0.75 M H3PO4 solution necessary to neutralize 35 mL of 0.45 M Ca(OH)2 is approximately 14 mL.
To determine the volume of 0.75 M H3PO4 necessary to neutralize 35 ml of 0.45 M Ca(OH)2, we can use the concept of stoichiometry.
Stoichiometry is the relationship between the relative quantities of substances taking part in a reaction. In this case, we can use the balanced equation to find the stoichiometric ratio between H3PO4 and Ca(OH)2.
From the balanced equation:
2 H3PO4(aq) + 3 Ca(OH)2 (aq) → Ca3(PO4)2 + 6 H2O
We can see that for every 2 moles of H3PO4, we need 3 moles of Ca(OH)2 to neutralize.
First, let's determine the number of moles of Ca(OH)2 in 35 mL of 0.45 M Ca(OH)2.
Number of moles = concentration (M) x volume (L)
0.45 M Ca(OH)2 x 0.035 L = 0.01575 moles Ca(OH)2
Now, using the stoichiometric ratio, we can determine the number of moles of H3PO4 needed to neutralize the given amount of Ca(OH)2.
Since the stoichiometric ratio is 2:3, the number of moles of H3PO4 required is:
(3/2) x 0.01575 moles Ca(OH)2 = 0.023625 moles H3PO4
Finally, let's find the volume of 0.75 M H3PO4 containing 0.023625 moles.
Volume (L) = moles / concentration (M)
Volume = 0.023625 moles / 0.75 M = 0.0315 L
Since the volume of the acid is typically expressed in milliliters, we can convert the volume to mL by multiplying by 1000.
Volume = 0.0315 L x 1000 mL/L = 31.5 mL
Therefore, the volume of 0.75 M H3PO4 necessary to neutralize 35 mL of 0.45 M Ca(OH)2 is 31.5 mL.