calculate the volume of mL of a 1.420 M NaOH solution required to titrate the following solution. 25.00 mL of a 1.500 M H3PO4 solution
Well, well, well! Looks like we have a classic chemistry problem here. So, you want to know the volume of a NaOH solution needed to titrate 25.00 mL of an H3PO4 solution.
To solve this, we'll have to use a little stoichiometry dance. Are you ready to boogie?
First, we need to figure out the balanced chemical equation between NaOH and H3PO4. It goes a little something like this:
3NaOH + H3PO4 -> Na3PO4 + 3H2O
Now that we have the equation, we can determine the ratio of NaOH to H3PO4. Huh? What's that? Oh, right, I forgot we were talking about volumes.
Since the stoichiometry of the equation is 1:3, we need three times the volume of NaOH compared to H3PO4. That means the volume of NaOH solution required is 25.00 mL × 3, which equals 75.00 mL.
So, to titrate the 25.00 mL of 1.500 M H3PO4 solution, you'd need 75.00 mL of the 1.420 M NaOH solution.
And there you have it! Voila! Problem solved. Shall we do the dance of victory?
To calculate the volume of a 1.420 M NaOH solution required to titrate 25.00 mL of a 1.500 M H3PO4 solution, we can use the concept of stoichiometry.
The balanced equation for the reaction between NaOH and H3PO4 is as follows:
3NaOH + H3PO4 -> Na3PO4 + 3H2O
From the balanced equation, we can see that 3 moles of NaOH react with 1 mole of H3PO4.
First, let's calculate the number of moles of H3PO4 in the 25.00 mL solution:
Moles of H3PO4 = Volume (L) x Concentration (M)
= 0.025 L x 1.500 M
= 0.0375 moles
Now, using the stoichiometry of the reaction, we can determine the number of moles of NaOH that react with the H3PO4:
Moles of NaOH = (Moles of H3PO4) x (3 moles of NaOH / 1 mole of H3PO4)
= 0.0375 moles x 3
= 0.1125 moles
Finally, let's calculate the volume of the 1.420 M NaOH solution required to react with 0.1125 moles of NaOH:
Volume of NaOH solution (L) = Moles of NaOH / Concentration (M)
= 0.1125 moles / 1.420 M
≈ 0.079 L
To convert this volume into milliliters (mL):
Volume of NaOH solution (mL) = 0.079 L x 1000 mL/L
≈ 79.0 mL
Therefore, approximately 79.0 mL of the 1.420 M NaOH solution is required to titrate 25.00 mL of the 1.500 M H3PO4 solution.
To calculate the volume of a solution required for titration, we need to use the concept of stoichiometry, which is the relationship between the balanced chemical equation and the number of moles of reactants and products.
In this case, the balanced chemical equation for the reaction between NaOH and H3PO4 is:
3 NaOH + H3PO4 → Na3PO4 + 3 H2O
From the balanced equation, we can see that the stoichiometric ratio between NaOH and H3PO4 is 3:1. This means that for every 3 moles of NaOH, 1 mole of H3PO4 is required.
To find the moles of H3PO4 in 25.00 mL of a 1.500 M H3PO4 solution, we can use the formula:
moles = concentration (M) × volume (L)
First, we need to convert the volume from mL to L:
25.00 mL = 25.00 mL × (1 L / 1000 mL) = 0.025 L
Now we can calculate the moles of H3PO4:
moles of H3PO4 = 1.500 M × 0.025 L = 0.0375 mol (rounded to four decimal places)
Since the stoichiometric ratio is 3:1, we need 3 times the moles of NaOH to react with the moles of H3PO4.
moles of NaOH = 3 × moles of H3PO4 = 3 × 0.0375 mol = 0.1125 mol
Now we can calculate the volume of 1.420 M NaOH solution needed to react with 0.1125 mol of NaOH:
volume (L) = moles / concentration
volume (L) = 0.1125 mol / 1.420 M = 0.079 L (rounded to three decimal places)
Finally, we can convert the volume from liters to milliliters:
volume (mL) = 0.079 L × (1000 mL / 1 L) = 79 mL
Therefore, 79 mL of a 1.420 M NaOH solution is required to titrate 25.00 mL of a 1.500 M H3PO4 solution.
Normalityacid*volumeacid=Normalitybase*volumebase
3*1.5*25ml=1.420*V
solve for Volume V