Question

31 formula 4 points
how many ml of a 0.16 m h₂so₄ solution will you need to neutralize 179 ml of 0.1 m naoh? solve to the hundredth place.
2 naoh(aq) + h₂so₄(aq) → 2 h₂o(l) + na₂so₄(aq)
answer
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Explanation:

Step1: Calculate the moles of NaOH

Use the formula \(n = M\times V\) (where \(n\) is moles, \(M\) is molarity, \(V\) is volume in liters).
\(V_{NaOH}=179\space mL = 0.179\space L\), \(M_{NaOH}=0.1\space M\)
\(n_{NaOH}=M_{NaOH}\times V_{NaOH}=0.1\times0.179 = 0.0179\space mol\)

Step2: Use the mole - ratio from the balanced equation

From \(2NaOH(aq)+H_2SO_4(aq)\to2H_2O(l)+Na_2SO_4(aq)\), the mole ratio \(n_{NaOH}:n_{H_2SO_4}=2:1\)
So \(n_{H_2SO_4}=\frac{n_{NaOH}}{2}=\frac{0.0179}{2}=0.00895\space mol\)

Step3: Calculate the volume of \(H_2SO_4\)

Use \(V=\frac{n}{M}\), \(M_{H_2SO_4} = 0.16\space M\), \(n_{H_2SO_4}=0.00895\space mol\)
\(V_{H_2SO_4}=\frac{n_{H_2SO_4}}{M_{H_2SO_4}}=\frac{0.00895}{0.16}=0.0559375\space L\)

Step4: Convert volume to mL

\(V_{H_2SO_4}=0.0559375\times1000 = 55.94\space mL\) (rounded to the hundredth place)

Answer:

\(55.94\space mL\)