Question

During a titration, an endpoint was reached after adding 20 mL of sodium hydroxide solution, NaOH(aq), to 20 mL of sulfuric acid, H2SO4. The sodium hydroxide solution had a concentration of 1.50 mol/L. What was the initial concentration of the sulfuric acid? Select one: a. 1.50 mol/L b. 2.25 mol/L c. 3 mol/L d. 0.75 mol/L

Answer 1

Write the balanced chemical equation for the reaction as follows:

`2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O`

Given:

Concentration of sodium hydroxide =1.50 mol/L

Volume of sodium hydroxide = 20 mL

Concentration of sulfuric acid = x

Volume of sulfuric acid = 20 mL

Firstly, we will determine the moles that is in 20 mL of 1.5 M NaOH:

`\begin{gathered} 1.5\text{ }mole=1000mL \\ x\text{ }mol=20mL \\ \\ x=(1.5mol*20mL)/(1000mL) \\ \\ x=0.03mol\text{ }NaOH \end{gathered}`

Based on the balanced chemical equation and the moles of NaOH we can use stoichiometry (ratios) to determine the moles of 20 mL sulfuric acid:

`0.03mol\text{ }NaOH*\frac{1mol\text{ }H_2SO_4}{2mol\text{ }NaOH}=0.015\text{ }mol`

Now that we know how many moles are in 20 mL of H2SO4, we can determine the moles in 1000mL (initial concentration):

`\begin{gathered} 0.015\text{ }mol=20\text{ }mL \\ x\text{ }mol=1000\text{ }mL \\ x=(0.015mol*1000mL)/(20mL) \\ \\ x=0.75mol•L^(-1) \end{gathered}`

Answer: D) 0.75 mol/L,