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? 0.22 mol/l 0.44 mol/l 0.066 mol/l 0.033 mol/l 0.11 mol/l

Explanation:

Step1: Write the balanced chemical equation

\[H_2SO_4 + 2NaOH
ightarrow Na_2SO_4+2H_2O\]
The mole - ratio of \(H_2SO_4\) to \(NaOH\) is \(n_{H_2SO_4}:n_{NaOH}=1:2\).

Step2: Calculate the moles of \(NaOH\)

Use the formula \(n = M\times V\), where \(M\) is the molarity and \(V\) is the volume in liters.
\(V_{NaOH}=50.0\ mL = 0.0500\ L\), \(M_{NaOH}=0.085\ mol/L\)
\(n_{NaOH}=M_{NaOH}\times V_{NaOH}=0.085\ mol/L\times0.0500\ L = 4.25\times 10^{-3}\ mol\)

Step3: Calculate the moles of \(H_2SO_4\)

From the mole - ratio, \(n_{H_2SO_4}=\frac{1}{2}n_{NaOH}\)
\(n_{H_2SO_4}=\frac{4.25\times 10^{-3}\ mol}{2}=2.125\times 10^{-3}\ mol\)

Step4: Calculate the molarity of \(H_2SO_4\)

\(V_{H_2SO_4}=19\ mL = 0.019\ L\)
Use the formula \(M=\frac{n}{V}\), so \(M_{H_2SO_4}=\frac{n_{H_2SO_4}}{V_{H_2SO_4}}\)
\(M_{H_2SO_4}=\frac{2.125\times 10^{-3}\ mol}{0.019\ L}=0.11\ mol/L\)

Answer:

0.11 mol/L