Question

In a precipitate reaction between BaCl2(aq) and Na3PO4(aq), 56.7 mL of 0.100 M BaCl2(aq) completely reacted with 25.3 mL of Na3PO4(aq). What was the molarity of Na3PO4(aq)? 3 BaCl2(aq) + 2 Na3PO4(aq) → Ba3(PO4)2(s) + 6 NaCl(aq) Note: Insert only the numerical value of your answer (do not include the units or chemical in your answer).

Answer 1

volume of BaCl2 used = 56.7ml , use following formula to find volume of BaCl2 used in liter

volume in liter = volume in ml/1000 , substitute values

volume of BaCl2 used = 56.7/1000 = 0.0567 liter BaCl2 used, molarity of BaCl2 used = 0.1 M

no of moles =molarity x volume in liter , substitute values in this formula to find no of moles of BaCl2 used

no of moles of BaCl2 used = 0.1 x 0.0567 = 0.00567 moles of BaCl2 used

balanced chemical reaction between BaCl2 and Na3PO4 is given bellow

3BaCl2 + 2Na₃PO₄ $\rightarrow$ Ba₃(PO₄)₂ + 6NaCl

according to balanced chemical reaction 3 moles of BaCl2 reacts with 2 moles of Na3PO4 that mean 3 moles of BaCl2 requre 2 moles of Na3PO4 to complete reaction therefore 0.00567 moles of BaCl2 will require

= 0.00567 x 2/3 = 0.00378 moles of Na3PO4 to complete given reaction

volume of Na3PO4 required = 25.3 ml = 25.3/1000 = 0.0253 liter no of moles of Na3PO4 required = 0.00378moles

molarity = no of moles/volume in liter , substitute values in this formula to find molarity of Na3PO4

molarity of Na3PO4 = 0.00378/0.0253 = 0.1494 M

ans molarity of Na3PO4_(aq) = 0.1494 M