A 25.0 mL sample of 0.25 M potassium carbonate (K2CO3) solution is added to 30.0 mL of a 0.40 M barium nitrate (Ba(NO3)2) solution. What is the concentration of the excess metal ion after the precipitation reaction is complete

0.10M of Ba²⁺ is the concentration of the metal in excess

Explanation:

Based on the chemical reaction:

K₂CO₃(aq) + Ba(NO₃)₂(aq) → BaCO₃(s) + 2KNO₃(aq)

1 mole of potassium carbonate reacts per mole of barium nitrate

To solve this question we need to find the moles of each salt to find then the moles of Barium in excess:

Moles K₂CO₃:

0.025L * (0.25mol / L) = 0.00625moles K₂CO₃ = moles CO₃²⁻

Moles Ba(NO₃)₂:

0.030L * (0.40mol/L) = 0.012 moles of Ba(NO₃)₂ = 0.012 moles of Ba²⁺

That means moles of Ba²⁺ that don't react are:

0.012 mol - 0.00625mol = 0.00575 moles Ba²⁺

In 25 + 30mL = 55mL:

0.00575 moles Ba²⁺ / 0.055L =