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To find the grams of H3PO4 in 265 ml of a 1.50 M solution, you first calculate the moles of H3PO4 using the molarity (1.50 mol/L) and volume (0.265 L), then convert moles to grams using the molar mass of H3PO4 (98.0 g/mol). Therefore, there would be approximately 39.2 grams of H3PO4 in 265 ml of the solution.
To conduct a titration using H3PO4 as the titrant, first prepare a solution of H3PO4 of known concentration. Then, add the H3PO4 solution to the analyte solution (the solution being titrated) using a burette until the endpoint is reached. The endpoint is typically indicated by a color change or a sudden change in pH. Record the volume of H3PO4 solution used at the endpoint to calculate the concentration of the analyte.
During the titration of H3PO4 with NaOH to determine the equivalence point, a known volume of H3PO4 is gradually added to a solution of NaOH until the reaction reaches a neutral point. This is detected using an indicator that changes color at the equivalence point. The volume of NaOH required to reach this point is used to calculate the concentration of H3PO4.
H3PO4 is an acid because it can donate a hydrogen ion (proton) when dissolved in water, resulting in the formation of hydronium ions. This gives the solution acidic properties, as it increases the concentration of H+ ions. Additionally, phosphoric acid (H3PO4) has a dissociation constant (Ka) which indicates its ability to donate hydrogen ions in solution.
To find the Ka of the first ionization of H3PO4, you need to set up an ICE table and use the formula for Ka. Given that the pH at equilibrium is 3, you can calculate the concentration of H+ ions at equilibrium. Then, use this concentration to calculate the concentration of the other species in the reaction and plug them into the Ka expression to find the Ka value.
To find the grams of H3PO4 in 265 ml of a 1.50 M solution, you first calculate the moles of H3PO4 using the molarity (1.50 mol/L) and volume (0.265 L), then convert moles to grams using the molar mass of H3PO4 (98.0 g/mol). Therefore, there would be approximately 39.2 grams of H3PO4 in 265 ml of the solution.
To conduct a titration using H3PO4 as the titrant, first prepare a solution of H3PO4 of known concentration. Then, add the H3PO4 solution to the analyte solution (the solution being titrated) using a burette until the endpoint is reached. The endpoint is typically indicated by a color change or a sudden change in pH. Record the volume of H3PO4 solution used at the endpoint to calculate the concentration of the analyte.
Calculus with (units) in brackets shows the way how and why.275 (g) / 97.9 (g/mol) = 2.81 mol in 120 L = 2.81 (mol) / 120 (L) = 2.34*10-2 (mol/L) = 0.0234 M H3PO4And since there are 3 protons available per mole: 0.0234 (M) H3PO4 * 3(N/M) = 0.0702 N
First, calculate the moles of water: 175g / 18.0116 g/mol = 9.721 moles. The total moles in the solution is the sum of water and phosphoric acid (H3PO4) moles. Next, calculate the mole fraction of H3PO4: moles of H3PO4 / total moles in the solution. This will give you the mole fraction of phosphoric acid in the solution.
During the titration of H3PO4 with NaOH to determine the equivalence point, a known volume of H3PO4 is gradually added to a solution of NaOH until the reaction reaches a neutral point. This is detected using an indicator that changes color at the equivalence point. The volume of NaOH required to reach this point is used to calculate the concentration of H3PO4.
H3PO4 is an acid because it can donate a hydrogen ion (proton) when dissolved in water, resulting in the formation of hydronium ions. This gives the solution acidic properties, as it increases the concentration of H+ ions. Additionally, phosphoric acid (H3PO4) has a dissociation constant (Ka) which indicates its ability to donate hydrogen ions in solution.
To find the Ka of the first ionization of H3PO4, you need to set up an ICE table and use the formula for Ka. Given that the pH at equilibrium is 3, you can calculate the concentration of H+ ions at equilibrium. Then, use this concentration to calculate the concentration of the other species in the reaction and plug them into the Ka expression to find the Ka value.
The balanced chemical equation for the reaction is: 3 H3PO4 + 2 NaOH -> Na3PO4 + 3 H2O From the equation, we can see that 3 moles of H3PO4 react with 2 moles of NaOH. Calculate the molar mass of H3PO4 and NaOH. Then, use the molar ratio from the balanced equation to calculate the amount of H3PO4 needed to react with 25.4g of NaOH.
To make a 1N solution of H3PO4 (phosphoric acid), dissolve 98 g of pure H3PO4 in water and dilute to 1 liter with water. Make sure to follow proper safety precautions when working with concentrated acids.
The chemical formula for phosphoric acid is H3PO4.
That depends entirely on the concentration of H3PO4 dissolved in the solvent. It can have many different values of pH ranging from below zero at high concentration to near 7 at very low concentration.
The pH of H3PO4 (phosphoric acid) depends on its concentration. For a 1M solution, the pH would be around 0.91. Since phosphoric acid is a weak acid that can donate three protons, its pH decreases with increasing concentration due to the dissociation of H+ ions.