(1)425ml of a saturated solution of lanthanum iodate, $L a {(I O_{3})}_{3}$ $La(IO_3)_3$ , has $2.93 \times 10^{- 4}$ $2.93 xx 10^-4$ mole of $L a^{3 +}$ $La^(3+)$ (a)what is the concentration of $I O_{3}^{-}$ $IO_3^-$ ? (b)calculate the solubility product of lantham iodate?

Question

(1)425ml of a saturated solution of lanthanum iodate, $L a {(I O_{3})}_{3}$ $La(IO_3)_3$ , has $2.93 \times 10^{- 4}$ $2.93 xx 10^-4$ mole of $L a^{3 +}$ $La^(3+)$ (a)what is the concentration of $I O_{3}^{-}$ $IO_3^-$ ? (b)calculate the solubility product of lantham iodate?

1 Answer

Impact of this question

5752 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-09T23:42:31

Here's what I got.

Explanation:

So, you know that you're dealing with a $425-mL$ $"425-mL"$ sample of a saturated solution of lanthanum(III) iodate, $La {({IO}_{3})}_{3}$ $"La"("IO"_3)_3$ .

You also know that this solution contains $2.93 \cdot 10^{- 4}$ $2.93 * 10^(-4)$ moles of lanthanum(III) cations, ${La}^{3 +}$ $"La"^(3+)$ .

Use this information to find the molarity of the lanthanum(III) cations - do not forget that molarity uses liters of solution!

$c = \frac{n}{V}$ $color(blue)(c = n/V)$

$[{La}^{3 +}] = \frac{2.93 \cdot 10^{- 4} moles}{425 \cdot 10^{- 3} L} = 6.89 \cdot 10^{- 4} M$ $["La"^(3+)] = (2.93 * 10^(-4)"moles")/(425 * 10^(-3)"L") = 6.89 * 10^(-4)"M"$

Now, the equilibrium reaction that describes the partial dissociation of lanthanum iodate in aqueous solution looks like this

$La {({IO}_{3})}_{3(s]} ⇌ {La}_{(aq]}^{3 +} + 3 {IO}_{3(aq]}^{-}$ $"La"("IO"_3)_text(3(s]) rightleftharpoons "La"_text((aq])^(3+) + color(red)(3)"IO"_text(3(aq])^(-)$

Notice that you have a $1 : 1$ $1:1$ mole ratio between lanthanum(III) iodate and the lanthanum(III) cations. This tells you that every mole of the solid that dissolves in solution will produce $1$ $1$ mole of lanthanum(III) cations.

Likewise, the $1 : 3$ $1:color(red)(3)$ mole ratio that exists between the solid and the iodate anions, ${IO}_{3}^{-}$ $"IO"_3^(-)$ , tells you that every mole of the solid that dissolves in solution will produce $3$ $color(red)(3)$ moles of iodate anions.

So, using this mole ratio, you can say that the molarity of the iodate anions will be $3$ $color(red)(3)$ times higher than that of the lanthanum(III) cations.

$[{IO}_{3}^{-}] = 3 \times [{La}^{3 +}]$ $["IO"_3^(-)] = color(red)(3) xx ["La"^(3+)]$

$[{IO}_{3}^{-}] = 3 \times 6.89 \cdot 10^{- 4} M = 2.07 \cdot 10^{- 3} M$ $["IO"_3^(-)] = color(red)(3) xx 6.89 * 10^(-4)"M" = color(green)(2.07 * 10^(-3)"M"$

The expression of the solubility product constant, $K_{s p}$ $K_(sp)$ , for lanthanum(III) iodate will look like this

$K_{s p} = [{La}^{3 +}] \cdot {[{IO}_{3}^{-}]}^{3}$ $K_(sp) = ["La"^(3+)] * ["IO"_3^(-)]^color(red)(3)$

Plug in your values to get

$K_{s p} = 6.89 \cdot 10^{- 4} \cdot {(2.07 \cdot 10^{- 3})}^{3}$ $K_(sp) = 6.89 * 10^(-4) * (2.07 * 10^(-3))^color(red)(3)$

$K_{s p} = 6.12 \cdot 10^{- 12}$ $K_(sp) = color(green)(6.12 * 10^(-12))$

The answers are rounded to three sig figs.

Science

Math

Humanities

... and beyond

(1)425ml of a saturated solution of lanthanum iodate, $L a {(I O_{3})}_{3}$ $La(IO_3)_3$ , has $2.93 \times 10^{- 4}$ $2.93 xx 10^-4$ mole of $L a^{3 +}$ $La^(3+)$ (a)what is the concentration of $I O_{3}^{-}$ $IO_3^-$ ? (b)calculate the solubility product of lantham iodate?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

(1)425ml of a saturated solution of lanthanum iodate, La(IO_3)_3La(IO3)3La(IO_3)_3, has 2.93 xx 10^-42.93×10−42.93 xx 10^-4 mole of La^(3+)La3+La^(3+) (a)what is the concentration of IO_3^-IO−3IO_3^-? (b)calculate the solubility product of lantham iodate?

1 Answer

Explanation:

Related questions

Impact of this question

(1)425ml of a saturated solution of lanthanum iodate, $L a {(I O_{3})}_{3}$ $La(IO_3)_3$ , has $2.93 \times 10^{- 4}$ $2.93 xx 10^-4$ mole of $L a^{3 +}$ $La^(3+)$ (a)what is the concentration of $I O_{3}^{-}$ $IO_3^-$ ? (b)calculate the solubility product of lantham iodate?