Answer on the question #43840, Chemistry, Other
Question:
What is the molarity of (SO4)^-2 ion in aqueous solution that contain 34.2ppm of Al2(SO4)3?
Solution:
If the solution contains the 34.2 ppm of Al2(SO4)3, it means that:
𝑐𝑐 =
𝑚𝑚(𝐴𝐴𝑙𝑙2 (𝑆𝑆𝑂𝑂4 )3 )
∗ 106 = 34.2 ppm
𝑚𝑚(𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠)
There is x grams if Al2(SO4)3in 100 g of solution:
𝑥𝑥 = 34.2 ∗ 100 ∗ 10−6 = 3.42 ∗ 10−3 𝑔𝑔
The number of moles of Al2(SO4)3 is:
m
10−3
= 10−5 𝑚𝑚𝑚𝑚𝑚𝑚
𝑛𝑛(𝐴𝐴𝑙𝑙2 (𝑆𝑆𝑂𝑂4 )3 ) = = 3.42 ∗
M
342.1509
The number of moles of Al2(SO4)3and of (SO4)2- ion relates as:
𝑛𝑛(𝑆𝑆𝑂𝑂4 2− ) = 3 ∗ 𝑛𝑛(𝐴𝐴𝑙𝑙2 (𝑆𝑆𝑂𝑂4 )3 ) = 3 ∗ 10−5 𝑚𝑚𝑚𝑚𝑚𝑚
Then, if the density of the solution is 1g/L, the molarity of (SO4)2- ion is:
𝑐𝑐(𝑆𝑆𝑂𝑂4
Answer: 3 ∗ 10−6 𝑚𝑚𝑚𝑚𝑚𝑚/𝐿𝐿
2−
n(𝑆𝑆𝑂𝑂4 2− )
10−5
)=
= 3∗
= 3 ∗ 10−6 𝑚𝑚𝑚𝑚𝑚𝑚/𝐿𝐿
V(solution)
0.1
http://www.AssignmentExpert.com/