Chemical and Ionic Equilibrium 2 Question 73
8. An aqueous solution contains an unknown concentration of $\mathrm{Ba}^{2+}$. When $50 \mathrm{~mL}$ of a $1 \mathrm{M}$ solution of $\mathrm{Na}{2} \mathrm{SO}{4}$ is added, $\mathrm{BaSO}{4}$ just begins to precipitate. The final volume is $500 \mathrm{~mL}$. The solubility product of $\mathrm{BaSO}{4}$ is $1 \times 10^{-10}$. What is the original concentration of $\mathrm{Ba}^{2+}$ ?
(2018 Main)
(a) $5 \times 10^{-9} \mathrm{M}$
(b) $2 \times 10^{-9} \mathrm{M}$
(c) $1.1 \times 10^{-9} \mathrm{M}$
(d) $1.0 \times 10^{-10} \mathrm{M}$
Show Answer
Answer:
Correct Answer: 8. (c)
| 9. $(\mathrm{d})$ | 10. $(\mathrm{b})$ | 11. (b) | 12. (d) |
|---|---|---|---|
| 13. $(\mathrm{d})$ | 14. (d) | 15. (b) | 16. (a) |
| 17. (c) | 18. (d) | 19. (a) | 20. (b) |
| 21. (d) | 22. (d) | 23. (a) | 24. (a) |
| 25. (d) | 26. (c) | 27. (a) | 28. (d) |
| 29. (c) | 30. (a) | 31. (b) | 32. (a) |
| 33. (a) | 34. (b) | 35. (d) | 36. (b) |
| 37. (c, d) | 38. $(\mathrm{a}, \mathrm{b}, \mathrm{c})$ | 39. (b, c) | 40. (4.47) |
| 41. $(\mathrm{d})$ | 42. $\mathrm{I}_{2}$ | 43. hydration | |
| 44. amphoteric | 45. $\mathrm{SO}_{4}^{2-}$ | 46. $\mathrm{F}$ | 47. $\mathrm{F}$ |
| 48. $\mathrm{F}$ | 49. $(3)$ | 50. $\left(1.6 \times 10^{-7}\right)$ | 52. $(8)$ |
| 53. $(9)$ | 55. $(4.86)$ | 56. $\left(1.2 \times 10^{-3} \mathrm{M}\right)$ | |
| 58. $\left(2 \times 10^{-8}\right)$ | 61. $(11.5)$ | 62. $(6.50)$ | 64. (1) |
| 65. $(80)$ | 67. $\left(1.8 \times 10^{-5}\right)$ | 68. $\left(9.67 \times 10^{-11}\right)$ | 69. |
| $\left(27.78 \times 10^{3}\right)$ | 74. $(4.20)$ | ||
| 71. $(0.177)$ | 72. $\left(8.7 \times 10^{-4} \mathrm{gL}^{-1}\right)$ | ||
| 75. $(99.83)$ | 77. $(>7)$ |
Solution:
- Its given that the final volume is $500 \mathrm{~mL}$ and this final volume was arrived when $50 \mathrm{~mL}$ of $1 \mathrm{M} \mathrm{Na}{2} \mathrm{SO}{4}$ was added to unknown $\mathrm{Ba}^{2+}$ solution.
So, we can interpret the volume of unknown $\mathrm{Ba}^{2+}$ solution as $450 \mathrm{~mL}$ i.e.
$$ \underset{\substack{\mathrm{Ba}^{2+} \ \text { solution }}}{450 \mathrm{~mL}}+\underset{\substack{\mathrm{Na}{2} \mathrm{SO}{4} \ \text { solution }}}{500 \mathrm{~mL}} \longrightarrow \underset{\substack{\mathrm{BaSO}_{4} \ \text { solution }}}{500 \mathrm{~mL}} $$
From this we can calculate the concentration of $\mathrm{SO}_{4}^{2-}$ ion in the solution via
$$ \begin{gathered} M_{1} V_{1}=M_{2} V_{2} \ 1 \times 50=M_{2} \times 500 \end{gathered} $$
(as $1 \mathrm{M} \mathrm{Na}{2} \mathrm{SO}{4}$ is taken into consideration)
$$ M_{2}=\frac{1}{10}=0.1 \mathrm{M} $$
Now for just precipitation,
$$ \begin{aligned} & \text { Ionic product }=\text { Solubility product }\left(K_{\mathrm{sp}}\right) \ & \text { i.e. } \quad\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{SO}{4}^{2-}\right]=K{\text {sp }} \text { of } \mathrm{BaSO}{4} \ & \text { Given } K{\text {sp }} \text { of } \mathrm{BaSO}_{4}=1 \times 10^{-10} \end{aligned} $$
$$ \begin{array}{rlrl} \text { So, } & & {\left[\mathrm{Ba}^{2+}\right][0.1]} & =1 \times 10^{-10} \ \text { or } & {\left[\mathrm{Ba}^{2+}\right]} & =1 \times 10^{-9} \mathrm{M} \end{array} $$
Remember This is the concentration of $\mathrm{Ba}^{2+}$ ions in final solution. Hence, for calculating the $\left[\mathrm{Ba}^{2+}\right]$ in original solution we have to use
$$ \begin{array}{rlrl} & M_{1} V_{1} & =M_{2} V_{2} \ & \text { as } & M_{1} \times 450 & =10^{-9} \times 500 \ \text { so, } & M_{1} & =1.1 \times 10^{-9} \mathrm{M} \end{array} $$
