Question

A scientist wants to make a solution of tribasic sodium phosphate, Na3PO4Na3PO4, for a laboratory experiment. How many grams of Na3PO4Na3PO4 will be needed to produce 625 mLmL of a solution that has a concentration of Na+Na+ ions of 1.00 MM ?

Answer 1

Answer:

102.5 grams of sodium phosphate will be needed.

Explanation:

$Na_3PO_4(aq)\rightarrow 3Na^+(aq)+PO_4^{3-}(aq)$

Molarity of the sodium ions = $[Na^+]=1.00 M$

$3\times[Na_3PO_4]= [Na^+]$

$[Na_2PO_4]=\frac{1}{3}\times 1.00 M=1.00 M$

Mass of the sodium phosphate = m

Molar mass of sodium phosphate = 164 g/mol

Volume of the sodium phosphate solution = 625 mL = 0.625 L

1 mL = 0.001 L

Molarity of the sodium phosphate = $[Na_3PO_4]=1.00 M$

$Molarity=\frac{\text{Mass of compound}}{\text{Molar mas of compound}\times Volume (L)}$

$1.00 M=\frac{m}{164 g/mol\times 0.625 L}$

m = 102.5 g

102.5 grams of sodium phosphate will be needed.