Calculate the volume of Ba(OH)2 0.20 M necessary to neutralize 300.0 mL of H3PO4 0.30 M. (4 marks)
how
To calculate the volume of Ba(OH)2 required to neutralize 300.0 mL of H3PO4, we can use the equation:
M1V1 = M2V2
where:
M1 = initial concentration of H3PO4 = 0.30 M
V1 = initial volume of H3PO4 = 300.0 mL
M2 = final concentration of Ba(OH)2 = 0.20 M
V2 = final volume of Ba(OH)2 (which we need to find)
Rearranging the equation, we have:
V2 = (M1 * V1) / M2
Now, plugging in the values:
V2 = (0.30 M * 300.0 mL) / 0.20 M
V2 = (0.09 mol/L * 0.300 L) / 0.20 mol/L
V2 = 0.03 L or 30.0 mL
Therefore, the volume of Ba(OH)2 0.20 M necessary to neutralize 300.0 mL of H3PO4 0.30 M is 30.0 mL.
To calculate the volume of Ba(OH)2 0.20 M necessary to neutralize 300.0 mL of H3PO4 0.30 M, we can use the concept of stoichiometry and the equation for the reaction between Ba(OH)2 and H3PO4.
The balanced chemical equation for the reaction is:
2 H3PO4 + 3 Ba(OH)2 → Ba3(PO4)2 + 6 H2O
From the balanced equation, we can see that it takes 3 moles of Ba(OH)2 to react with 2 moles of H3PO4.
Now let's calculate the moles of H3PO4 using its molarity and volume:
moles of H3PO4 = molarity of H3PO4 × volume of H3PO4
= 0.30 M × 0.300 L
= 0.09 moles
Since the stoichiometric ratio is 2:3 for Ba(OH)2:H3PO4, we can use this ratio to calculate the moles of Ba(OH)2 required:
moles of Ba(OH)2 = (moles of H3PO4 × 3) / 2
= (0.09 moles × 3) / 2
= 0.135 moles
Now, let's calculate the volume of Ba(OH)2 using its molarity and the moles calculated:
volume of Ba(OH)2 = moles of Ba(OH)2 / molarity of Ba(OH)2
= 0.135 moles / 0.20 M
= 0.675 L
= 675 mL
Therefore, the volume of Ba(OH)2 0.20 M necessary to neutralize 300.0 mL of H3PO4 0.30 M is 675 mL.
3Ba(OH)2 + 2H3PO4 ==> Ba3(PO4)2 + 6H2O
mols H3PO4 = M x L = ?
Use the coefficients to convert from mols H3PO4 to mols Ba(OH)2. Then, M Ba(OH)2 = mols Ba(OH)2/L Ba(OH)2