What are the respective concentrations (moles/liter) of k+ and po43- afforded by dissolving 0.800 mol k3po4 in water and diluting to 1.63l?

Given Data:

   No. of moles of K3PO4 = 0.800 mole

  Molarity of K3PO4 =  0.800÷1.63 = 0.4908

Chemical equations:

        K3PO4 ⇔3 K+   +  PO4∧-2

Solution:

         K3PO4 : K+                                                K3PO4 : PO4∧-2

            1  :    3                                                            1      :    1

        0.4908 = 0.4908 × 3 = 1.47 M                   0.4908 = 0.4908 × 1 = 0.4908 M

the answer is d. 74.02

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explanation:

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