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Which of the following salt will produce a neutral solution (If any)? A) $\ce{KI}$ (Potassium Iodine) or B) $\ce{SrS}$ (Strontium Sulfur)

I tried to solve this problem by using the "Relative Strengths of Brønsted-Lowry Acids and Base in aq solution at room temp table". From what I know each of those two substance will dissociate into ions. Having the ions $\ce{K^+, I^-,Sr^{2+}, S^{2-}}$

Since K does not hydrolyze and I is a product of HI which is a strong acid it won't hydrolyze either. I also know that the only cations that hydrolyze are Al,Fe and Cr hence Sr will not hydrolyze. However how do I know weather S will hydrolyze? The other question I had difficulties was "Order the following from most acidic to least. SO3, CO3 and CO2". enter image description here

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In my homework one of the question was which of the following salt will produce a neutral solution(If any)? A) KI (Potassium Iodine) or B) SrS (Strontium Sulfur)

For (A) both K+ and I- will essentially stay as the ions. So assuming the solution is neutral to start, it will remain neutral.

But for (B), in a neutral solution, $\ce{Sr^{2+}}$ will stay as the ion, $\ce{S^{2-}}$ will react as:

$\ce{S^{2-} + H2O <=> HS^- + OH-}$

so the solution will become slightly more basic.

"Order the following from most acidic to least. $\ce{SO3}$, $\ce{CO3}$ and $\ce{CO2}$".

I can't really make sense of this question since $\ce{CO3}$ is highly unstable, and there are three isomers.

Bubbling $\ce{SO3}$ and $\ce{CO2}$ gases into water would produce acidic solutioons via reactions

$\ce{SO3 + H2O -> H2SO4}$

$\ce{CO2 + H2O -> H2CO3}$

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  • $\begingroup$ Just to clarify in the second question we evaluate the acidity by the acid that is produce as the final product not the acids that are formed as the intermediates such as HSO4 and HCO3? $\endgroup$ – coderhk Jul 21 '17 at 22:33
  • $\begingroup$ There are really no such species as $\ce{HSO4}$ and $\ce{HCO3}$ in aqueous solution. The species would have to be $\ce{HSO4^-}$ and $\ce{HCO3^-}$ which would formed from acid reacting with $\ce{H2O}$. $\ce{H2SO4}$ is such a strong acid that there would be just a trace of $\ce{HSO4^-}$, with most of the S species being $\ce{SO4^{2-}}$. $\endgroup$ – MaxW Jul 21 '17 at 23:16

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