Estimate the pH of the solution that results when this added to 50 mL of 0.275 M Na2HPO4(aq)

- 50 mL of 0.275 M HCl(aq)

millimols HPO4^- = mL x M = approx 14 but you need to use a better number than that so recalculate all of these.

mmols HCl added = 50 x 0.275 = approx 14.

.......HPO4^2- + H^+ ==> H2PO4^-
I......14........0........0
add .............14............
C.....-14.......-14......+14
E......0..........0.......14

So you have prepared 14 mmols H2PO4^- in 100 mL so the pH will be determined by the 0.14M acid (that's 0.275/2 = ?).
......H2PO4^- + H2O ==> OH^- + H3PO4
I.....0.14...............0.......0
C......-x................x.......x
E.....0.14-x.............x.......x

Kb for H2PO4^- = (Kw/Ka for H3PO4) = (x)(x)/(0.14-x)
Solve for x = (OH^-) and convert to pH. Post your work if you get stuck.

To determine the pH of the resulting solution when HCl is added to Na2HPO4, we need to consider the balanced chemical equation for the reaction between the two:

2 Na2HPO4 + 2 HCl → 2 NaCl + H2O + H3PO4

Since the reaction involves a 1:1 ratio of HCl to Na2HPO4, all of the HCl will be used up to react with the Na2HPO4.

First, we need to calculate the number of moles of HCl used:

moles of HCl = concentration (M) × volume (L)
= 0.275 M × 0.05 L
= 0.01375 moles

Next, we can calculate the moles of Na2HPO4 used based on the stoichiometry of the balanced equation. From the equation, we can see that 2 moles of HCl react with 2 moles of Na2HPO4. Therefore, the number of moles of Na2HPO4 used will be half of the moles of HCl used:

moles of Na2HPO4 used = 0.01375 moles ÷ 2
= 0.006875 moles

To find the final concentration of H3PO4 in the solution, we can divide the moles of H3PO4 by the total volume of the solution (50 mL + 50 mL = 100 mL = 0.1 L):

final concentration of H3PO4 = moles of H3PO4 ÷ volume of solution
= 0.006875 moles ÷ 0.1 L
= 0.06875 M

Since H3PO4 is a weak acid, it partially dissociates in water. In this case, we have a diluted solution, so we can assume that the dissociation is complete. The ionization equation for H3PO4 is:

H3PO4 ⇌ H+ + H2PO4-

The pH of the solution can be calculated using the concentration of H+ (acidic concentration). In our case, the concentration of H+ will be equal to the final concentration of H3PO4:

pH = -log10 [H+]
= -log10 (0.06875)
= 1.162

Therefore, the estimate of the pH of the resulting solution is approximately 1.162.

To estimate the pH of the resulting solution, we need to consider the reaction that occurs when Na2HPO4 and HCl are mixed together. In this case, Na2HPO4 is a base and HCl is an acid.

The reaction between Na2HPO4 and HCl is as follows:

2 Na2HPO4(aq) + 2 HCl(aq) → 2 NaCl(aq) + H2PO4-(aq) + H2O(l)

The balanced equation shows that for every 2 moles of Na2HPO4 and HCl, we get 1 mole of H2PO4- ions.

To calculate the concentration of H2PO4-(aq) in the resulting solution, we need to determine the moles of H2PO4- and the volume of the final solution.

First, let's calculate the moles of H2PO4- generated:

Moles of H2PO4- = (50 mL of 0.275 M Na2HPO4 / 1000 mL/L) x (1 L/1000 mL) x (2 mol H2PO4-/1 mol Na2HPO4) = 0.0275 mol H2PO4-

Since the reaction is a 1:1 ratio between Na2HPO4 and H2PO4-, we know that the concentration of H2PO4- is also 0.0275 M.

Next, we need to calculate the total volume of the final solution:

Total volume = 50 mL + 50 mL = 100 mL = 0.1 L

Now we have the moles of H2PO4- and the volume of the solution. To estimate the pH, we can use the Henderson-Hasselbalch equation:

pH = pKa + log([A-] / [HA])

In this case, H2PO4- acts as a weak acid and its conjugate base is HPO4^2-. The pKa of the H2PO4-/HPO4^2- system is around 7.2 at room temperature. Therefore, we can use this value to estimate the pH of the solution.

pH = 7.2 + log(0.0275 M / 0.0275 M) = 7.2

Therefore, the estimated pH of the resulting solution is 7.2.