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12 April, 20:28

A scientist wants to make a solution of tribasic sodium phosphate, Na3PO4, for a laboratory experiment. How many grams of Na3PO4 will be needed to produce 525 mL of a solution that has a concentration of Na ions of 1.10 M

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  1. 12 April, 20:42
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    31.652g of Na3PO4

    Explanation:

    We'll begin by calculating the molarity of Na3PO4 solution. This can be achieved as shown below:

    Na3PO4 will dessicate in solution as follow:

    Na3PO4 (aq) - > 3Na + (aq) + PO4³¯ (aq)

    From the balanced equation above,

    1 mole of Na3PO4 produce 3 moles of sodium ion, Na+.

    Therefore, xM Na3PO4 will produce 1.10M sodium ion, Na + i. e

    xM Na3PO4 = (1.10 x 1) / 3

    xM Na3PO4 = 0.367M

    Therefore, the molarity of Na3PO4 is 0.367M.

    Next, we shall determine the number of mole of Na3PO4 in the solution. This is illustrated below:

    Molarity of Na3PO4 = 0.367M

    Volume = 525mL = 525/1000 = 0.525L

    Mole of Na3PO4 = ... ?

    Molarity = mole / Volume

    0.367 = mole / 0.525

    Cross multiply

    Mole of Na3PO4 = 0.367 x 0.525

    Mole of Na3PO4 = 0.193 mole.

    Finally, we shall convert 0.193 mole of Na3PO4 to grams. This is illustrated below:

    Molar mass of Na3PO4 = (23x3) + 31 + (16x4) = 164g/mol

    Mole of Na3PO4 = 0.193 mole

    Mass of Na3PO4 =.?

    Mass = mole x molar mass

    Mass of Na3PO4 = 0.193 x 164

    Mass of Na3PO4 = 31.652g

    Therefore, 31.652g of Na3PO4 is needed to prepare the solution.
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