The Ka values of H3PO4 are shown below.

Ka1 7.5 10-3
Ka2 6.2 10-8
Ka3 4.8 10-13

What is the pH of a 0.19 M solution of NaH2PO4?

(H^+) = sqrt(k1k2)

How come it didn't matter what the molarity was? It wasn't used at all.

same question. why wasn't the molarity used>?

Well, first things first, let's clown around and calculate the concentration of H+ ions in the solution. Sodium phosphate, NaH2PO4, can be considered a salt of a weak acid (H3PO4). So, we should expect some acidic action going on.

Given that the concentration of NaH2PO4 is 0.19 M, we can assume that it fully dissociates into H+ and H2PO4- ions in solution. Since sodium ion, Na+, is a spectator, we can happily ignore it (spectators don't like being ignored, but what can you do?).

Now, let's keep this clown car going and write an equation for the dissociation of H2PO4-:

H2PO4- ⇌ H+ + HPO42-

The Ka2 value is given as 6.2x10-8, which is the equilibrium constant for this reaction. Since this is a weak acid, we can assume that the initial concentration of H+ ions is negligible compared to the final concentration, so we can let x be the concentration of H+, which is also the concentration of H2PO4-.

Now, what does the equilibrium expression look like for this clowny affair? It's the Ka2 expression, of course:

Ka2 = [H+][HPO42-] / [H2PO4-]

Plugging in what we know, we get:

6.2x10-8 = [x][x] / (0.19 - x)

Let's say x is so small compared to 0.19 that it might as well be zero (as clowns often do). This allows us to simplify the equation to:

6.2x10-8 = x^2 / 0.19

Now, let's solve for x (don't worry, I have my trusty clown calculator with me!). By rearranging the equation, we get:

x^2 = 0.19 * 6.2x10-8
x^2 = 1.178x10-8
x ≈ 3.43x10-5

Okay, almost there, clown comrades! Now that we have the concentration of H+ ions, let's calculate the pH. The pH is defined as the negative logarithm (base 10) of the H+ concentration, so we have:

pH ≈ -log(3.43x10-5)
pH ≈ 4.46

So, the pH of the 0.19 M solution of NaH2PO4 is approximately 4.46. Let's juggle some acids and bases now!

To find the pH of a solution of NaH2PO4, we need to consider the acid-base properties of H3PO4, as NaH2PO4 is the sodium salt of H3PO4.

H3PO4 is a triprotic acid, meaning it can donate three protons (H+ ions) in solution. Each protonation step has its own equilibrium constant (Ka value). The three Ka values given represent the dissociation constants for each protonation step.

The three protonation steps for H3PO4 are:
H3PO4 ⇌ H+ + H2PO4- (Ka1)
H2PO4- ⇌ H+ + HPO42- (Ka2)
HPO42- ⇌ H+ + PO43- (Ka3)

NaH2PO4 is the sodium salt of H2PO4-. It is a salt composed of a weak acid (H2PO4-) and the conjugate base of a strong acid (Na+). In aqueous solution, NaH2PO4 will partially dissociate to produce H2PO4- ions and Na+ ions.

Since H2PO4- is an acidic species, it can act as a weak acid and donate protons in solution. The equilibrium constant for this dissociation is Ka2.

To calculate the pH of a 0.19 M solution of NaH2PO4, we need to find the concentration of H+ ions produced by the dissociation of H2PO4-.

Step 1: Set up an ICE table for the dissociation reaction:
H2PO4- ⇌ H+ + HPO42-
Initial Concentration (M): 0.19
Change in Concentration (M): -x + x
Equilibrium Concentration (M): 0.19 - x

Step 2: Write the expression for the equilibrium constant (Ka2):
Ka2 = [H+][HPO42-]/[H2PO4-]

Since we assume that x is small compared to 0.19, we can approximate the equilibrium concentration of H2PO4- as 0.19 M.

Ka2 = [H+][HPO42-]/(0.19)

Step 3: Substitute the value of Ka2 and solve for x:
6.2 x 10^-8 = x^2/(0.19)
x^2 = (0.19)(6.2 x 10^-8)
x^2 = 1.178 x 10^-8
x = √(1.178 x 10^-8)
x ≈ 3.43 x 10^-5

Step 4: Calculate the pH of the solution:
pH = -log[H+]
pH = -log(3.43 x 10^-5)
pH ≈ 4.465

Therefore, the pH of a 0.19 M solution of NaH2PO4 is approximately 4.465.