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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Win,odd Dhamnekar on June 10, 2021, 10:52:26 AM

In all the following exercises M represents metal and X represents anion.
a) The salt, MX_{2} has a K_{sp} of 2.00e6. If equal volumes of 0.0059M M(NO_{3})_{2} and 0.01M NaX are mixed, will solid form?
b)The salt, MX_{3} has a K_{sp} of 2.07e10. If equal volumes of 0.0029M M(NO_{3})_{3} and 0.02M NaX are mixed, will solid form?
c)The salt, MX has a K_{sp} of 2.13e6. If equal volumes of 0.0062M M(NO_{3})_{2} and 0.03M Na_{2}X are mixed, will solid form?
My worked out answers:
a)Q_{ion}=0.00295*0.01^{2}=2.95e7 which is less than the given K_{sp} 2.00e6, hence solid won't form.
b)Q_{ion}=0.00145*(0.03)^{3}=3.915e8 which is greater than the given K_{sp} 2.07e10, hence, solid will form.
c)Q_{ion}=0.0031*0.015= 465e5 which is greater than the given K_{sp} 2.13e6, hence solid will form.
What is your opinion about these answers? Are these answer correct? In my opinion, all these three answers are correct.

Where do you get the value 0.03 in b? Or 0.01 in a?

In b) , Initial concentration of salt MX_{3} is 0.02M.After mixing equal volumes of metal(M) and anion(X), it will be 0.01. Then it will be multiplied by X's subscript in MX_{3}=0.03.
Similar is the case for (a).

You don't have MX_{3}, you have M(NO_{3})_{3} and NaX. After mixing, you have 0.00145 M M^{3+} and 0.01 M X^{}.
You mustn't assume there are stiochiometric amounts of M and X!