How do you prepare one liter of phosphate buffer 0.01 M at pH 12.0 from 85% phosphoric acid (density 1.69g/ml) and NaOH 1.0 M
A fairly long solution.
Start with two equations. The first is to find the ratio of base to acid.
pH = pK3 + log (base)/(acid)
Substitute and solve for b/a.
Equation 2 is b + a = 0.01.
Solve those two equations simultaneously for a (acid) and b (base). Since those are M concns and you want a L those will be mols in a L. I estimate mols acid = about 0.007 and mols base = about 0.003
How do you get 0.007 mols H3PO4. That will be 0.007 (remember to do these calculations more accurately than my estimates) x 98 g/mol = approx 0.7 g.
How to get approx 0.7 grams H3PO4.
mL x density x %acid = 0.7 and solve for mL H3PO4.
For the base, you want approx 0.003 mols. M = mols/L. YOu know M and mols, solve for L and convert to mL if needed.
There is another way to do this and the problem may be asking for that solution.
Calculate the a and b as above, then calculate mL H3PO4. Place in a beaker, add water, stir, insert pH meter electrodes into the solution and add 1.0 M NaOH until the pH reads 12.0. Dilute to 1 L.
Post your work if you get stuck.
To prepare one liter of phosphate buffer 0.01 M at pH 12.0, you would need to utilize the properties of phosphoric acid (H3PO4) and sodium hydroxide (NaOH). Here's a step-by-step guide to help you:
1. Determine the amount of phosphoric acid needed:
- Since the density of the 85% phosphoric acid is given as 1.69 g/ml, we need to find the volume of phosphoric acid required to achieve the desired concentration.
- The molar mass of H3PO4 is 97.99 g/mol (1 x 1.01 g/mol for hydrogen, 3 x 16.00 g/mol for oxygen, and 1 x 31.00 g/mol for phosphorus).
- The desired concentration is 0.01 M, which means there should be 0.01 moles of H3PO4 in 1 liter of solution.
- Convert moles to grams: 0.01 moles x 97.99 g/mol = 0.9799 grams
Note: Since the phosphoric acid in the question is 85%, it means that 100 grams of the solution contains 85 grams of H3PO4. We can use this information to calculate the volume.
- Calculate the volume of phosphoric acid needed: 0.9799 grams x (100 g / 85 g) = 1.1534 grams
2. Calculate the volume of sodium hydroxide needed:
- The NaOH concentration given is 1.0 M, so we need to find the number of moles of NaOH required to neutralize the phosphoric acid.
- The neutralization reaction between phosphoric acid and sodium hydroxide is as follows:
H3PO4 + 3 NaOH → Na3PO4 + 3 H2O
- From the balanced equation, we see that 1 mole of H3PO4 reacts with 3 moles of NaOH.
- Since we need 0.01 moles of H3PO4, we will have 0.01 moles x (3 moles of NaOH / 1 mole of H3PO4) = 0.03 moles of NaOH.
Note: Molarity (M) is defined as moles of solute per liter of solution. Therefore, we need 0.03 moles of NaOH in 1 liter of solution.
- Convert moles to grams: 0.03 moles x 39.997 g/mol (molar mass of NaOH) = 1.19991 grams
3. Determine the final volume of the buffer solution:
- Since we need to prepare a 1.0 L solution, the final volume will be 1.0 L.
4. Calculate the remaining volume of water needed:
- To obtain a final volume of 1.0 L, we need to subtract the volumes of the phosphoric acid and sodium hydroxide from the total volume.
- Remaining volume of water = 1.0 L - volume of phosphoric acid - volume of sodium hydroxide
Note: The volume of a solution can be calculated using its mass and density, where Volume (V) = Mass (m) / Density (D).
- Volume of phosphoric acid = 1.1534 grams / 1.69 g/ml (density of phosphoric acid) = 0.682 ml
- Volume of sodium hydroxide = 1.19991 grams / 1.0 g/ml (assumed density of sodium hydroxide) = 1.19991 ml
- Remaining volume of water = 1.0 L - 0.682 ml - 1.19991 ml = 997.11809 ml
5. Prepare the buffer solution:
- Take a 1.0 L container and add the calculated volumes of phosphoric acid (0.682 ml), sodium hydroxide (1.19991 ml), and water (997.11809 ml).
- Mix the solution thoroughly to ensure proper dissolution and homogeneity.
By following these steps, you will be able to prepare one liter of phosphate buffer 0.01 M at pH 12.0 using 85% phosphoric acid and NaOH 1.0 M.