A student dilutes a solution of 0.100 mol of H3PO4 and 0.150 mol of NaOH with...

A student dilutes a solution of 0.100 mol of H3PO4 and 0.150 mol of NaOH with water to a volume of 1.00 L but then realizes that the "distilled water" used was in fact dilute hydrochloric acid.

(A) If the pH of the final solution is 6.62, what are the concentrations of H3PO4, H2PO^−4, and HPO^2−4?

(B) How many moles of HCl were inadvertently added to the original solution?

Many Thanks

Expert Solution

In preparing the original solution, the NaOH will neutralize the first H+ of H3PO4 completely, and will neutralize 1/2 of the H2PO4-, so that the solution contains 0.05 M H2PO4- and 0.05 M HPO42-. The pKa's for the protons of phosphoric acid are:
pKa1 = 2.12
pKa2 = 7.21
pKa3 = 12.38

Only the second pKa is in play here. Now, using that pKa2, you can calculate the actual final concentrations using the Henderson-Hasselbalch equation:
pH = pKa + log [HPO42-]/[HPO4-]
6.62 = 7.21 + log [HPO42-]/[H2PO4-]
[HPO42-]/[H2PO4-] = 0.257

Now, since [HPO42-] + [H2PO4-] = 0.100 M and [HPO42-] = 0.257[H2PO4-],
0.257[H2PO4-] + [H2PO4-] = 0.100
1.257[H2PO4-] = 0.100
[H2PO4-] = 0.0796 M
[HPO42-] = 0.0204 M

You can use the expression for Ka1 to calculate [H3PO4] in the solution:
Ka1 = [H+][H2PO4-]/[H3PO4]
7.5X10^-3 = 3.16X10^-7(0.0796)/[H3PO4]
[H3PO4] = 3.35X10^-6 M

b) In the absence of added HCl, [HPO42-] = [H2PO4-] = 0.05 M
So, because you are dealing with 1 L of solution, moles of each would have equaled 0.050 mol.
In the actual final solution, moles H2PO4-] = 0.0796, so moles HCl added = 0.0796 - 0.0500 = 0.0296

queen_honey_blossom answered 2 years ago

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