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chemistrynerd12 · Oct 16, 2009, 10:42 PM

Question: You have 0.59g of KH2PO4 diluted to 250mL. What is the concentration of phosphate?

I calculated the concentration of the solution, which was 0.01735 M, but there asking for the concentration of just the phosphate? Confused.

Do you take the grams and divide by just the number of K, H and O?
Is the concentration of phosphate the same as the solution?

Perito · Oct 17, 2009, 05:58 AM

Question: You have 0.59g of KH2PO4 diluted to 250mL. What is the concentration of phosphate?

I calculated the concentration of the solution, which was 0.01735 M, but there asking for the concentration of just the phosphate? Confused.

do you take the grams and divide by just the number of K, H and O?
Is the concentration of phosphate the same as the solution?

You started with 0.59 grams of KH2PO4. The molecular weight of KH2PO4 is 136.086 g/mole. You, therefore, have

$\frac {0.59\, g} {136.086 \frac {g}{mole}}\, = \, 0.004335\, moles\, of \, KH_2PO_4$

The concentration of KH2PO4 is 0.004553 moles/0.25 L = 0.01734 m/L

which is what you got within a decimal point.

Phosphate is simply the ion $PO_4^{3-}$ .

You have the following equilibrium:

$KH_2PO_4 \rightleftharpoons K^+ + 2H^+ + PO_4^{3-}$

For one molecule of KH2PO4, you get 1 potassium ion, two hydrogen ions, and one phosphate ion. Therefore, the concentration of potassium is 0.01734 m/L. The concentration of hydrogen is 2 x 0.01734 m/L = 0.0347 m/L, and the concentration of phosphate is 0.01734 m/L.

In this case, the concentration of phosphate is the same as the solution because one molecule of KH2PO4 yields one phosphate ion.

chemistrynerd12 · Oct 17, 2009, 01:44 PM

Beautiful! Thank-you so much. Getting that incorrect would've ruined my whole lab report. I really appreciate it

:)