Question

Xianming runs a titration and collects, dries, and weighs the BaSO4(s) produced in the experiment. He reports a mass of 0.2989 g go BaSO4. Based on this, calculate the concentration of Ba(OH)2 solution.

Answer 1

The question is incomplete, here is the complete question:

$Ba(OH)_2(aq.)+H_2SO_4(aq.)\rightarrow BaSO_4(s)+2H_2O(l)$

Xianming runs a titration and collects, dries, and weighs the $BaSO_4(s)$ produced in the experiment. He reports a mass of 0.2989 g go $BaSO_4$ Based on this, calculate the concentration of $Ba(OH)_2$ solution.

Answer: The concentration of barium hydroxide solution is 0.0013 moles.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$  

Given mass of barium sulfate = 0.2989 g

Molar mass of barium sulfate = 47.87 g/mol

Putting values in above equation, we get:

$\text{Moles of barium sulfate}=\frac{0.2989g}{233.4g/mol}=0.0013mol$

The given chemical reaction follows:

$Ba(OH)_2(aq.)+H_2SO_4(aq.)\rightarrow BaSO_4(s)+2H_2O(l)$

By Stoichiometry of the reaction:

1 mole of barium sulfate is produced from 1 mole of barium hydroxide

So, 0.0013 moles of barium sulfate will be produced from = $\frac{1}{1}\times 0.0013=0.0013mol$ of barium hydroxide

As, no volume of the container is given. So, the concentration will be calculated in moles only.

Hence, the concentration of barium hydroxide solution is 0.0013 moles.