a solution made by dissolving 2.733g KH2PO4 and 4.508g K2HPO4 in water to give 250 ml of solution.
is the answer...
7.432= ph
What's the question?
What is the pH of the buffer solution
The base is K2HPO4. The acid is KH2PO4.
Convert g base to mols. mols = grams/molar mass. Do the same for the acid. Convert each to M. M = mols/L.
Then substitute into the HH equation and solve for pH.
thank you very much, do i used the molar mass of KH2PO4 altogether or just of H2PO4?
You use the molar mass of the material for which you have a mass; i.e., if your material weighed is KH2PO4 then mols KH2PO4 = grams KH2PO4/molar mass kH2PO4.
To determine the pH of a solution made by dissolving KH2PO4 and K2HPO4 in water, we need to consider the dissociation of the phosphate salts in water.
First, let's find the moles of each compound:
Moles of KH2PO4 = mass / molar mass = 2.733g / (39.10g/mol + 2 * 1.01g/mol + 32.00g/mol + 16.00g/mol) = 0.0159 mol
Moles of K2HPO4 = mass / molar mass = 4.508g / (2 * 39.10g/mol + 32.00g/mol + 4 * 1.01g/mol) = 0.0297 mol
Now, let's calculate the total moles of phosphate ions (HPO4^2-) present in the solution:
Total moles of HPO4^2- = moles of KH2PO4 + 2 * moles of K2HPO4 = 0.0159 mol + 2 * 0.0297 mol = 0.0753 mol
In a solution made from the combination of KH2PO4 and K2HPO4, the dominant species in terms of phosphate ions is HPO4^2-. Therefore, we can consider HPO4^2- as a weak acid.
To determine the pH, we need to know the dissociation constant (Ka) of HPO4^2-:
Ka = [H3O+][HPO4^2-] / [H2PO4^-]
The expression for Ka is not given, so we can't directly calculate the pH. However, if we assume the dissociation of HPO4^2- is complete, we can calculate the concentration of the resulting H3O+ ions.
Based on this assumption, the concentration of H3O+ ions can be calculated using the stoichiometry of the balanced equation for the dissociation of HPO4^2-:
HPO4^2- + H2O ⇌ H2PO4^- + OH-
Since the concentration of OH- is equal to the concentration of H3O+, we can calculate it. Here, we'll assume that the initial concentration of HPO4^2- is equal to the total moles of HPO4^2- divided by the solution volume in liters:
[HPO4^2-] = total moles of HPO4^2- / solution volume in liters = 0.0753 mol / (250 mL / 1000 mL/L) = 0.301 mol/L
[H2PO4^-] = [OH-] = [H3O+]
Now, we'll use the expression for the ion product of water (Kw) to find the concentration of H3O+ ions:
Kw = [H3O+][OH-] = 10^-14
[H3O+]^2 = 10^-14
[H3O+] = sqrt(10^-14) = 10^-7 mol/L
Finally, we can calculate the pH using the formula:
pH = -log[H3O+]
pH = -log(10^-7) = 7
Therefore, the pH of the solution will be 7.