Question

The concentration of so42–(aq) in a sample of river water can be determined using a precipitation titration in which a salt of ba2+(aq) is the standard solution and baso4(s) is the precipitate formed. what is the concentration of so42– (aq) in a 56.1 ml sample if 5.00 ml of a 0.00100 m ba2+(aq) solution is needed to precipitate all the so42–(aq) in the sample?

Answer 1

Hello!

The chemical reaction of the titration is the following:

Ba⁺²(aq) + SO₄⁻²(aq) → BaSO₄(s)

To calculate the concentration, we need to know the moles of SO₄⁻² in the initial solution, in the following way:

$molesSO_4^{-2}=[Ba^{+2} ]*VBa^{+2}(L)=0,001M*0,005 L \\ \\ molesSO_4^{-2}=0,000005 moles$

Now we divide the moles between the volume of the sample:

$[SO_4^{-2}]= \frac{molesSO_4^{-2}}{V sol (L)}= \frac{0,000005 molesSO_4^{-2}}{0,0561 L}=0,000089M$

So the concentration of SO₄⁻² is 0,000089M

Have a nice day!

Answer 2

Answer: The concentration of sulfate ions in the solution is $8.91\times 10^{-5}M$

Explanation:

To calculate the number of moles for given molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$        .......(1)

For barium ions:

Molarity of solution = 0.00100 M

Volume of solution = 5.00 mL

Putting values in equation 1, we get:

$0.00100M=\frac{\text{Moles of barium ions}\times 1000}{5.00}\\\\\text{Moles of barium ions}=\frac{0.00100\times 5.00}{1000}=5\times 10^{-6}moles$

The chemical equation for the ionization of barium sulfate follows:

$BaSO_4\rightarrow Ba^{2+}+SO_4^{2-}$

By Stoichiometry of the reaction:

1 mole of barium ions precipitate 1 mole of sulfate ions

So, $5\times 10^{-6}$ moles of barium ions will precipitate $\frac{1}{1}\times 5\times 10^{-6}=5\times 10^{-6}$ moles of sulfate ions

Now, calculating the molarity of sulfate ions by using equation 1:

Moles of sulfate ions = $5\times 10^{-6}$ moles

Volume of solution = 56.1 mL

Putting values in equation 1, we get:

$\text{Molarity of sulfate ions}=\frac{5\times 10^{-6}\times 1000}{56.1}\\\\\text{Molarity of sulfate ions}=8.91\times 10^{-5}M$

Hence, the concentration of sulfate ions in the solution is $8.91\times 10^{-5}M$