Chapter 15, Problem 8STP

Interpretation Introduction

Interpretation:

The ion left in the reaction after completion of the reaction needs to be determined.

Concept introduction:

When barium nitrate is reacted with potassium phosphate, barium phosphate and potassium nitrate is produced.

Answer to Problem 8STP

D

Explanation of Solution

Given:

Molarity of $Ba{\left(N{O}_3\right)}_2$ = $0.1M$

Volume of $Ba{\left(N{O}_3\right)}_2$ = $250mL$

Molarity of $K_{3}P{O}_4$ = $0.1M$

Volume of $K_{3}P{O}_4$ = $200mL$

Calculations:

Moles of $Ba{\left(N{O}_3\right)}_2$ reacted= $0.1\times \frac{250}{1000}$

= $0.025M$

Molar mass of $Ba{\left(N{O}_3\right)}_2$ =137+2(14+48)

=261 g/mol

Mass of $Ba{\left(N{O}_3\right)}_2$ = $0.025\times 261$

=6.525g

Moles of $K_{3}P{O}_4$ reacted= $0.1\times \frac{200}{1000}$

= $0.020M$

Molar mass of $K_{3}P{O}_4$ =117+31+64

=212 g/mol

Mass of $K_{3}P{O}_4$ = $0.020\times 212$

=4.24 g

The reaction is given as:

  $Ba{\left(N{O}_3\right)}_2+K_{3}P{O}_4\to B{a}_3{\left(P{O}_4\right)}_2+KN{O}_3$

According to reaction:

261 g $Ba{\left(N{O}_3\right)}_2$ reacts with 212 g $K_{3}P{O}_4$

So, 6.525 g $Ba{\left(N{O}_3\right)}_2$ should react with = $6.525\times \frac{212}{261}$

=5.3 g

But $K_{3}P{O}_4$ is available in lesser amount. So, it is limiting reagent and reaction will be based on it.

Whole of $K_{3}P{O}_4$ will react but some amount of $Ba{\left(N{O}_3\right)}_2$ will be left unreacted.

So, after reaction the ions present should be barium nitrate and phosphate.

Conclusion

Mathematical calculation shows that option D is correct and rest options are wrong.