The concentration of #"IO"_3^(-)# ions in a pure, saturated solution of #"Ba"("IO"_3)_2# is #1.06xx10^(-3)"M"#. What is the #K_(sp)# for #"Ba"("IO"_3)_2#?

Answer 1

#K_(sp) = 5.96 * 10^(-10)#

The problem tells you that a saturated solution of barium iodate, #"Ba"("IO"_3)_2#, has a concentration of iodate anions, #"IO"_3^(-)#, equal to #1.06 * 10^(-3)"M"#.
This means that when barium iodate is dissolved in water, the molar concentration of the dissociated iodate anions will be equal to #1.06 * 10^(-3)"M"#.

The dissociation equilibrium for barium iodate in aqueous solution looks like this

#"Ba"("NO"_ 3)_ (color(red)(2)(s)) rightleftharpoons "Ba"_ ((aq))^(2+) + color(red)(2)"IO"_ (3(aq))^(-)#
Notice that every mole of barium iodate that dissociates produces #1# mole of barium cations, #"Ba"^(2+)#, and #color(red)(2)# moles of iodate anions.

This means that in a saturated barium iodate solution, the concentration of iodate anions will be twice as high as the concentration of barium cations.

This means that the latter will be equal to

#["Ba"^(2+)] = 1/color(red)(2) * ["IO"_3^(-)]#
#["Ba"^(2+)] = 1/color(red)(2) * 1.06 * 10^(-3)"M" = 5.30 * 10^(-4)"M"#
Now, the solubility product constant, #K_(sp)#, for this dissociation equilibrium is defined as
#K_(sp) = ["Ba"^(2+)] * ["IO"_3^(-)]^color(red)(2)#

Plug in your values to find

#K_(sp) = 5.30 * 10^(-4)"M" * (1.06 * 10^(-3)"M")^color(red)(2)#
#K_(sp) = 5.96 * 10^(-10)"M"^3#

Solubility product constants are usually given without added units, which means that you answer is

#K_(sp) = color(green)(|bar(ul(color(white)(a/a)color(black)(5.96 * 10^(-10))color(white)(a/a)|)))#

The answer is rounded to three sig figs.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

[ K_{\text{sp}} = [Ba^{2+}][IO_3^-]^2 ]

[ K_{\text{sp}} = (4 \times 1.06 \times 10^{-3})^2 ]

[ K_{\text{sp}} = 4.49 \times 10^{-6} ]

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7