Determine the pH of 0.250 M sodium phosphate, Na 3 PO 4 . Phosphoric acid, H 3 PO 4 , contains K a1 = 7.5 × 10 − 3 , K a2 = 6.2 × 10 − 8 and K a3 = 4.2 × 10 − 13 .

Determine the pH of a 0.250 molar sodium phosphate solution. Here we're told that phosphoric acid itself possesses 3 ka values. If we were to closely examine sodium phosphate you would see that it's connected to 3 sodium ions, but more importantly it's connected to a phosphate ion. The phosphate ion here represents the basic form of our triprotic acid. This is the form after it's given away all 3 of its acidic H+ ions. And remember, for the basic form we utilize K b 1. And if since it's made by giving away the last acidic hydrogen, k b 1 is connected to k a 3, and together their product gives me k w. Here we're gonna say kvone equalskw divided by ka 3, so that's 1.0 times 10 to the negative 14, divided by 4.2 times 10 to the negative 13. Doing this gives us a k b one equal to 2.38 times 10 to the negative 2. Since k b one is less than 1, we know that this is a weak base. Alright, so we're gonna bring down the phosphate ion, it being a base it reacts with water and it accepts an H+ from the water, making it hydrogen phosphate ion plus the hydronium ion as products. It's a weak base, so we're gonna have to utilize an ICE chart to find our pH. Remember, in an ice chart we ignore solids and liquids, so water would be ignored. The initial concentration of our base is 0.250, products here initially are 0. We lose reactants to make product. We bring down everything, With our eyes chart filled out we can finally set up our equilibrium expression. So, KB 1 equals products over reactants, and with that we can start plugging in our values. KB1 is 2.38 times 10 to the negative two equals x squared divided by 0.250 minus x. This minus x we can examine to see can I ignore it? And we do that by doing our 500 approximation method. So we take our initial concentration and we divide it by k b 1. If this ratio comes out greater than 500 we can ignore the minus x. So we have 0.250 divided by 2.38 times 10 to the negative 2. Unfortunately, that only comes up to 10.5, which means we have to keep that minus x and solve using the quadratic formula. Cross multiply, so distribute, distribute. So we'd have 0.00595 minus 2.38 times 10 to the minus 2x equals x squared. Rearranging this equation where we bring everything to the right side we would add 2.38 times 10 to the negative 2x to both sides and subtract 0.00595 from both sides which would give me my new equation of x squared, plus 2.38 times 10 to the negative 2 x, minus 0.00595. This would be connected to a, b, and c, And with it I can use the quadratic formula. So we're gonna have negative 2.38 times 10 to the negative 2, plus or minus square root of 2.38 times 10 to the negative 2 squared minus 4 times 1 times negative 0.00595 divided by 2 times 1. We're gonna solve for everything here. When I do that I'm gonna get negative 2.38 times ten to the minus 2, plus or minus 0.1561 divided by 2. Because of this plus or minus I get 2 possible answers for x. X, if I added 0.1561, would give me x equals 0.06615 molar, and if I did minus 0.1561, x would equal negative 0.08995 molar. We can't utilize this negative x because plugging it into our equilibrium row of the ice chart would give us back a negative answer. Plus again here or here will give us back a negative answer, so I can't utilize it. That means we're gonna go with the positive x value. This is important because that positive x is connected to Oh minus, And if I know Oh minus concentration, then I know pOH, because pOH equals negative log of Oh minus. So let's plug it in. POH would equal 1 point 18. And if I know pOH then I know pH by default, because pH equals 14 minus pOH. So 14 minuteus 1.18 gives me 12.82 as the pH for this weak base.

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17. Acid and Base Equilibrium

Triprotic Acids and Bases Calculations

General Chemistry

17. Acid and Base Equilibrium

Triprotic Acids and Bases Calculations

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Multiple Choice

Determine the pH of 0.250 M sodium phosphate, Na₃PO₄. Phosphoric acid, H₃PO₄, contains K_a1 = 7.5 × 10⁻³, K_a2 = 6.2 × 10⁻⁸ and K_a3 = 4.2 × 10⁻¹³.

12.82

7.21

1.18

12.38

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Verified step by step guidance

Identify the relevant chemical species: Sodium phosphate (Na3PO4) is a salt derived from the weak acid phosphoric acid (H3PO4). In solution, Na3PO4 dissociates completely into 3 Na+ ions and one PO4^3- ion.

Recognize that PO4^3- is the conjugate base of HPO4^2-, which is the conjugate base of H2PO4^-, which in turn is the conjugate base of H3PO4. Therefore, PO4^3- can accept protons in a stepwise manner, and we need to consider the hydrolysis of PO4^3- in water.

Use the equilibrium expression for the hydrolysis of PO4^3-: PO4^3- + H2O ⇌ HPO4^2- + OH-. The equilibrium constant for this reaction can be derived from the relationship Kw = Ka3 * Kb, where Kw is the ion-product constant of water (1.0 × 10^-14 at 25°C).

Calculate the Kb for PO4^3- using the relationship Kb = Kw / Ka3. Substitute the known values: Kb = (1.0 × 10^-14) / (4.2 × 10^-13).

Set up the expression for the equilibrium concentration of OH- using the Kb expression: Kb = [HPO4^2-][OH-] / [PO4^3-]. Assume that the initial concentration of PO4^3- is 0.250 M and that the change in concentration is x, then solve for x, which represents [OH-]. Finally, calculate the pOH and convert it to pH using the relationship pH + pOH = 14.