b)What is the pH in the correctly ordered second scene (Ka1=7.2X10^-3, Ka2=6.3X1

Question

b)What is the pH in the correctly ordered second scene (Ka1=7.2X10^-3, Ka2=6.3X10^-8)?

pH=?

c) If it requires 18.00 mL of the NaOH solution to reach this scene, how much more is needed to reach the last scene?

mL NaOH?

The scenes below depict the relative concentrations of H3PO4, H2PO4 , and HPO during a titration with aqueous NaOH, but they are out of order. (Phosphate groups are purple, hydrogens are blue, and Na* ions and water molecules are not shown.) 8 l pause (a) List the scenes in the correct order. First Last

Explanation / Answer

a) List the scenes in the correct order. First Last

Note that initially, we will expect plenty of H3PO4, then as we add more NaOH, expect neutralization of protons so

H3PO4 > H2PO4- > HPO4-2 > PO4-3

From the list; expect:

C = a mixture of H3PO4 alone

B = a mixture of H3PO4 and H2PO4-

D = a mixture ofr H2PO4- and HPO4-2

E = only HPO4- present

b)What is the pH in the correctly ordered second scene (Ka1=7.2X10^-3, Ka2=6.3X10^-8)?

pH=?

First, identify the "second scene"

this is "C"

there is a mix of H3PO4 and H2PO4-

note that this is a buffer, by definition

there is acid, H3PO4 and conjugate base H2PO4-

so

pH = pKa + log(ratio)

ratio = H2PO4- / H3PO4

pKa = -log(Ka) = -log(7.2*10^-3) = 2.14266

now, ratio is = no of H2PO4- / no of H3PO4 = 3/3 = 1

pH = 2.14266 + log(1)

pH = 2.14 approx

c) If it requires 18.00 mL of the NaOH solution to reach this scene, how much more is needed to reach the last scene?

mL NaOH?

last scene is complete neutralization of H3PO4 to HPO4-2

if 18 mL were needed to half neutralize H3PO4 to H2PO4-

then, 36 mL will be neded to neutralize all H3PO4 to H2PO4-

further more, another 36 mL required to neutralize H2PO4- to HPO4-2

so

V total = 36 + 36 = 72 mL required to achieve last scene

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