Question

With all the steps please!What is the concentration of a sulfuric acid solution if 100 mL of the solution is neutralized by 50 mL of 0.5M Ba(OH)2 solution?

Answer 1

In this case we have a neutralization reaction, the balanced equation of the reaction is the following:

$Ba(OH)_{2(aq)}+H_2SO_{4(aq)}\rightarrow BaSO_{4(s)}+2H_2O_{(l)}$

Now, to find the concentration of the sulfuric acid solution we will follow the following steps:

1. We calculate the moles of Ba(OH)2 present in the solution. For this we use the definition of molarity that tells us:

$Molarity=\frac{MolesSolute}{Lsolution}$

2. We calculate the moles of H2SO4 that neutralize the calculated moles of Ba(OH)2. We find it by stoichiometry.

3. We calculate the concentration of the solution by applying the same molarity expression as above.

Let's proceed with the calculation:

1. Moles of Ba(OH)2

$\begin{gathered} MolesSolute=Lsolution\times Molarity \\ MolesBa(OH)_2=50mL\times\frac{1L}{1000mL}\times0.5M \\ MolesBa(OH)_2=0.025molBa(OH)_2 \end{gathered}$

2. Moles of H2SO4

By stoichiometry, we see that the H2SO ti Ba(OH)2 ratio is 1/1. So, the moles of H2SO4 will be the same of Ba(OH)2 moles

$\begin{gathered} MolH_2SO_4=GivenmolBa(OH)_2\times\frac{1molH_2SO_4}{1molBa(OH)_2} \\ MolH_2SO_4=0.025molBa(OH)_2\times\frac{1molH_2SO_4}{1molBa(OH)_2}=0.025molH_2SO_4 \end{gathered}$

3. Concentration of the solution

$\begin{gathered} Molarity=\frac{MolesSolute}{Lsolution} \\ Molar\imaginaryI ty=\frac{0.025molH_2SO_4}{100mL\times\frac{1L}{1000mL}} \\ Molar\imaginaryI ty=\frac{0.025molH_2SO_4}{0.1L}=0.25M \end{gathered}$

Answer: the concentration of H2SO4 solution is 0.25M