Question

A mixture is prepared by adding 18.3 mL of 0.183 M Na₃PO₄ to 30.3 mL of 0.153 M Ca(NO₃)₂.

What will be the concentrations (M) of each of the ions in the mixture after reaction? Put your answers with 3 significant digits.

A. NO₃^-

B. Na⁺

C. PO₄^3-

Answer 1

the balanced equation for the reaction :

2Na3PO4 + 3Ca(NO3)2 ---> 6NaNO3 + Ca3(PO4)2

calculate the number of moles of each reactant

Na3PO4 : 18.3 ml ( 0.183 mole / 1000 ml ) = 0.00335 moles

Ca(NO3)2 : 30.3 ml ( 0.153 mole / 1000 ml) = 0.00463 moles

(3/2) × 0.00335 = 0.005025 moles of Ca(NO3)2 has been consumed during the reaction .

Therefore 0.005025 - 0.00463 = 0.000395 moles have remained.

The final volume of the solution is 48.6 ml

therefore the concentrations of

[Na+]= (0.00335)/(48.6×10^-3) = 0.0689 M

There's actually no (PO4)3- remained as ion in the solution. so [PO4 3-] = 0 M

[NO3-]= (0.00463)/(48.6×10^-3) = 0.0952 M