Question

question 17 (1 point)
if it takes 19 ml of h₂so₄(aq) to neutralize 50.0 ml of 0.085 mol/l naoh(aq) then what is the molar concentration of the h₂so₄ solution?
0.11 mol/l
0.033 mol/l
0.22 mol/l
0.066 mol/l
0.44 mol/l

Explanation:

Step1: Write the balanced chemical equation

$$\ce{H_{2}SO_{4}(aq) + 2NaOH(aq) ightarrow Na_{2}SO_{4}(aq) + 2H_{2}O(l)}$$

From the equation, the mole ratio \(n(\ce{H_{2}SO_{4}}):n(\ce{NaOH}) = 1:2\)

Step2: Calculate the moles of \(\ce{NaOH}\)

Use the formula \(n = c\times V\) (where \(c\) is concentration and \(V\) is volume in liters).
\(V(\ce{NaOH})=50.0\space mL=50.0\times10^{- 3}\space L\), \(c(\ce{NaOH}) = 0.085\space mol/L\)
\(n(\ce{NaOH})=c(\ce{NaOH})\times V(\ce{NaOH})=0.085\space mol/L\times50.0\times 10^{-3}\space L = 4.25\times10^{-3}\space mol\)

Step3: Calculate the moles of \(\ce{H_{2}SO_{4}}\)

Since \(n(\ce{H_{2}SO_{4}})=\frac{1}{2}n(\ce{NaOH})\) (from mole - ratio)
\(n(\ce{H_{2}SO_{4}})=\frac{1}{2}\times4.25\times 10^{-3}\space mol=2.125\times10^{-3}\space mol\)

Step4: Calculate the concentration of \(\ce{H_{2}SO_{4}}\)

\(V(\ce{H_{2}SO_{4}})=19\space mL = 19\times10^{-3}\space L\)
Use \(c=\frac{n}{V}\), \(c(\ce{H_{2}SO_{4}})=\frac{2.125\times10^{-3}\space mol}{19\times10^{-3}\space L}\approx0.11\space mol/L\)

Answer:

0.11 mol/L