Chemistry, 23.03.2020 19:56 niescarlosj

A saturated solution of Cu(IO3)2 was prepared in 0.0250 M KIO3. The copper concentration was found to be 0.00072 M. Set up an ICE chart and calculate the equilibrium (IO3-) and the Ksp

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Answer from: tsdean21

0.02644 M = [IO3-]

Ksp = 0.0000005 = 5*10^-7

Explanation:

Step 1: Data given

Asaturated solution of Cu(IO3)2 was prepared in KIO3

Concentration of KIO3 = 0.0250 M

The copper concentration was found to be 0.00072 M

Step 2: Equations

Cu(IO3)2(s) + aq ⇔Cu2+(aq) + 2IO3 - (aq)

Step 3: The initial concentrations

The initial concentration of KIO3 = 0.0250 M

The initial concentration of Cu2+ is 0 M

Step 4: The changed concentrations

After preparing the Cu2+ in KIO3-; the concentration of Cu2+ will be changed with X

For every 1 Cu2+ we have 2 IO3-

This means if the concentration of Cu2+ is changed by X; the concentration of IO3- is changed by 2X. This gives a total concentration of 0.0250 + 2X for IO3-

Step 4:

Said is that the copper concentration was found to be 0.00072 M

This means the final concentration of Cu2+ is 0.00072 M or also said X = 0.00072

Now we can calculate the final IO3- concentration

0.0250 + 2X = [IO3-]

0.0250 + 2* 0.00072 = [IO3-]

0.0250 + 0.00144 = [IO3-]

0.02644 M = [IO3-]

Step 5: Calculate Ksp

Cu(IO3)2(s) ⇔Cu2++ 2IO3 -

Ksp of Cu(IO3)2 = [Cu+2]*[IO3-]²

Ksp = 0.00072 * 0.02644²

Ksp = 0.0000005 = 5*10^-7

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Answer from: BustD0wnAnt

$\large \boxed{\text{0.0264 mol/L; }5.0 \times 10^{-7}}$

Explanation:

1. The ICE table

$\begin{array}{rccccc} &\text{Cu}\text{(IO}_{3})_{2} & \rightleftharpoons &\text{Cu}^{2+}&+ & 2\text{IO}_{3}^{-}\\\\\text{I:}& & & 0 & & 0.0250\\\text{C:}& & & +x & & x \\\text{E:}& & & x & &0.0250 + 2x\\\end{array}$

2. Concentration of IO₃⁻

At equilibrium, [Cu²⁺] = 0.000 72 mol·L⁻¹, so x = 0.000 72.

The new ICE table becomes

$\begin{array}{rccccc} &\text{Cu}\text{(IO}_{3})_{2} & \rightleftharpoons &\text{Cu}^{2+}&+ & 2\text{IO}_{3}^{-}\\\\\text{I:}& & & 0 & & 0.0250\\\text{C:}& & & +0.00072 & & 0.00144 \\\text{E:}& & & 0.00072 & & \mathbf{0.0264}\\\end{array}\\\text{The equilibrium concentration of iodate ion is $\large \boxed{\textbf{0.0264 mol/L}}$}$

3. Ksp

$K_{\text{sp}} = [\text{Cu}^{2+}][\text{IO}_{3}^{-}]^{2} = 0.00072 \times 0.0264^{2} = \mathbf{5.0 \times 10^{-7}}$

A saturated solution of Cu(IO3)2 was prepared in 0.0250 M KIO3. The copper concentration was found t

Answer from: Quest

false

because it does not have the potential to be entirely used up

Answer from: Quest

"among the transition metal elements, manganese and zinc have a slight lowering of electronegativities. the reason behind this is that the nuclear charges of these elements are slightly weaker than the other transition metal elements due to the way their valence electrons are filled up. both manganese and zinc have their valence electrons filled up to the 4th shell with 1 pair of electrons (4s2) completely occupying it. although the electrons in the 3rd shell (3d) are yet to be paired since the outer most shell is filled up, the ability of the nucleus to attract electrons is lessened. hence, the electronegativity (or the ability to attract electrons) is lessened."

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A saturated solution of Cu(IO3)2 was prepared in 0.0250 M KIO3. The copper concentration was found t...