I want to replace 5g per liter of nah2po4 .h2o with h3po4 at 85%. the liquid is 1.685 g/ml
very simple replace it
i ned to know how what volume of liquid H3PO4 at 85% will be the equivalent to 5g of NaH2PO4. and how this calculated.
thanks
To replace 5g of NaH2PO4·H2O with H3PO4 at 85% concentration and with a liquid density of 1.685 g/mL, we need to calculate the volume required for the desired amount of H3PO4.
Step 1: Calculate the molar mass of NaH2PO4·H2O.
The molar mass of NaH2PO4·H2O can be calculated as follows:
- Molar mass of Na (22.99 g/mol) + Molar mass of H2PO4 (97.99 g/mol) + Molar mass of H2O (18.02 g/mol)
= 22.99 + 97.99 + 18.02
= 138 g/mol
Step 2: Calculate the moles of NaH2PO4·H2O.
To calculate moles, we use the formula:
Moles = Mass / Molar mass
Moles = 5 g / 138 g/mol
Moles ≈ 0.036 moles
Step 3: Convert moles of NaH2PO4·H2O to moles of H3PO4.
Since the ratio of NaH2PO4·H2O to H3PO4 is 1:1, the moles of H3PO4 will be the same as the moles of NaH2PO4·H2O.
Therefore, the moles of H3PO4 will also be approximately 0.036 moles.
Step 4: Calculate the mass of H3PO4 using the given concentration.
The mass of H3PO4 can be calculated as follows:
Mass = Concentration * Volume
5g = 0.85 * Volume
Volume ≈ 5g / 0.85
Volume ≈ 5.88 mL
Therefore, to replace 5g of NaH2PO4·H2O with H3PO4 at 85% concentration, you will need approximately 5.88 mL of H3PO4.