Determine the pH of a solution that is generated when 2.623 g Na3PO4 is added to a 100mL of 0.0800 M H3PO4.
I would use
k1k2k3 = (H^+)^3(PO4^3-)/(H3PO4)
Plug in for PO4^3- and H3PO4 and solve for (H^+) then convert to pH.
I use the Given concentrations? and solve for [H3O]?
right. mols Na3PO4 = grams/molar mass and M = mols/L solution.
H3PO4 is given.
Solve for H^+ and convert to pH.
To determine the pH of the solution, you need to consider the dissociation reactions of both Na3PO4 and H3PO4.
First, let's start with Na3PO4. Na3PO4 is a salt composed of sodium ions (Na+) and phosphate ions (PO4^3-). When Na3PO4 dissolves in water, it dissociates completely into its ions:
Na3PO4 -> 3Na+ + PO4^3-
Since sodium ions do not act as acids or bases in water, we can ignore them in the pH calculation. Therefore, we will focus on the phosphate ion (PO4^3-).
Next, let's consider H3PO4. H3PO4 is a weak acid and will partially dissociate in water:
H3PO4 + H2O -> H2PO4- + H3O+
This reaction establishes an equilibrium, and the concentration of H3O+ ions determines the pH of the solution.
Now, let's calculate the amount of H3PO4 in moles. We know that the volume of the H3PO4 solution is 100 mL, and the concentration of H3PO4 is 0.0800 M. Therefore, the number of moles of H3PO4 is:
moles H3PO4 = concentration H3PO4 x volume H3PO4 solution
= 0.0800 mol/L x 0.100 L
= 0.00800 mol
Next, we need to determine the concentration of the H3O+ ions at equilibrium. Since H3PO4 is a weak acid, we can assume that most of it remains undissociated. Therefore, the concentration of H3O+ ions will be approximately equal to the concentration of H3PO4 initially added.
Now, let's calculate the concentration of the H3O+ ions:
concentration H3O+ = concentration H3PO4
= 0.00800 M
Finally, to determine the pH, we need to calculate the negative logarithm (base 10) of the H3O+ concentration:
pH = -log[H3O+]
= -log(0.00800)
= 2.10
Therefore, the pH of the solution generated when 2.623 g Na3PO4 is added to 100 mL of 0.0800 M H3PO4 is approximately 2.10.