Solve Polyprotic pH Problem: K2HPO4 & H3PO4 (65 Chars)

[Zealduke]

A .05 M sample of K2HPO4 was dissolved in 800 mL of water, and the pH was adjusted to 7 by using a 1 M solution of H3PO4 followed by water bringing the volume up to 1 L. How many mL of H3PO4 were added? (On this problem charge and mass balance have to be used.)

My problem is just in the very beginning, as my book is terrible, I am not entirely sure on how to approach this problem.

Im assuming the mass balance for H3PO4 will be:

[H3PO4] + [H2PO4- ] + [HPO42-] + [PO43-] = 1 M

And the charge balance for K2HPO4 would be:

[H+] + [K+] = [OH-] + [H2PO4-] + 2*[HPO42-] + 3*[PO43-]


I'm really uncertain whether I'm missing anything or what the next step should be.

 

[Borek]

Please try these lectures on pH calcualtion as a starting point.

At pH 7.0 you can ignore presence of H₃PO₄ and PO₄^3-. I have a gut feeling that this question can be solved just using Henderson-Hasselbalch equation and stoichiometry.

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[Blueshift5]

I would appreciate any guidance or suggestions.Hello,

Thank you for sharing your approach to the problem. It seems like you have a good understanding of mass and charge balance, which are important principles to use in solving this type of problem.

One way to approach this problem is to set up an equation based on the information given. We know that the initial concentration of K2HPO4 is 0.05 M and the final volume is 1 L, so we can write:

(0.05 M)(800 mL) = (x M)(y mL)

where x is the concentration of H3PO4 and y is the volume of H3PO4 added. We can then use this equation to solve for y, which will give us the volume of H3PO4 added. This approach takes into account both the mass and charge balance, as the initial and final concentrations of K2HPO4 must be equal.

Another approach could be to use the Henderson-Hasselbalch equation, which relates the pH of a solution to the pKa and concentrations of the acid and its conjugate base. In this case, we know the final pH is 7 and the pKa for H3PO4 is 2.15, so we can use the equation:

pH = pKa + log([HPO42-]/[H3PO4])

to solve for the concentration of H3PO4, which will also give us the volume added.

I hope this helps and gives you some ideas on how to approach this problem. Keep in mind that there may be more than one way to solve it, so don't be afraid to try different approaches and see what works best for you. Good luck!