The solubility of two slightly soluble salts of M2 + , MA and MZ2, is the same, 4 * 10-4 mol/L. (c) If you added an equal volume of a solution saturated in MA to one saturated in MZ2, what would be the equilibrium concentration of the cation, M2+?

Video transcript

Hi everyone this problem reads A. X. And A. Y. Two R. Two slightly soluble salts of the A. I. On both of the salts have the same soluble itty of 6.2 times 10 to the negative five moles per liter. When a saturated solution of A. X. Is added to a saturated solution of A. Y. Two. What is the final equilibrium concentration of the A ion in solution? Okay. So we want to know the final equilibrium concentration of the ai on in the solution. All right, so let's look at what do we have initially? Initially we have, the concentration of the A ion is equal to 6.2 times 10 to the negative five molar. Okay. After the addition of A X. What's going to happen to the moles after the addition of A. X. The moles of the A. I. On is going to increase. Okay. So if we add A. X. This leads to the moles of our A. I. On increasing but the volume is also going to increase. Okay, so what's happening here is the concentration of the A ion remains the same. Okay, the concentration remains the same. The moles is increasing and the volume is increasing but the concentration remains the same. Okay, so that's it for this problem. I hope this was helpful