# The Ksp for BaCO3 is 5.1*10^-9. How many grams of BaCO2 will dissolve in 1000 ml of water?

Te idea here is that you need to use an ICE table to determine barium carbonate's molar solubility, then use the compound's molar mass to determine how many grams will dissolve in that much water.

This means that the molar solubility of barium carbonate will be equal to

I'll leave the answer rounded to two sig figs, despite the fact that you only have one sig figs for the volume of the water.

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The first step is to write the dissociation equation for BaCO3:

BaCO3 (s) ↔ Ba^2+ (aq) + CO3^2- (aq)

The Ksp expression for this equilibrium is:

Ksp = [Ba^2+] * [CO3^2-]

Given that the Ksp for BaCO3 is 5.1 × 10^-9, we can assume that x moles of BaCO3 will dissolve to give x moles of Ba^2+ and x moles of CO3^2-.

So, the equilibrium expression becomes:

5.1 × 10^-9 = x * x

Now, solve for x:

x^2 = 5.1 × 10^-9

x = sqrt(5.1 × 10^-9)

x ≈ 7.14 × 10^-5 moles/L

Since we want to find the grams of BaCO3 that will dissolve in 1000 mL of water (which is 1 L), we multiply the concentration by the volume:

( \text{Moles of BaCO}_3 = 7.14 \times 10^{-5} , \text{mol/L} )

( \text{Volume of solution} = 1000 , \text{mL} = 1 , \text{L} )

( \text{Mass of BaCO}_3 = (\text{Moles of BaCO}_3) \times (\text{Molar mass of BaCO}_3) )

The molar mass of BaCO3 is:

( \text{Molar mass of BaCO}_3 = 137.33 , \text{g/mol} + 12.01 , \text{g/mol} + (3 \times 16.00 , \text{g/mol}) )

( \text{Molar mass of BaCO}_3 = 137.33 , \text{g/mol} + 12.01 , \text{g/mol} + 48.00 , \text{g/mol} )

( \text{Molar mass of BaCO}_3 = 197.34 , \text{g/mol} )

Now, calculate the mass:

( \text{Mass of BaCO}_3 = (7.14 \times 10^{-5} , \text{mol/L}) \times (197.34 , \text{g/mol}) )

( \text{Mass of BaCO}_3 = 0.014 , \text{g} )

So, approximately 0.014 grams of BaCO3 will dissolve in 1000 mL of water.

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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.

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