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Chapter 4: Q4.52P (page 180)

If 26.25 mL of a standard 0.1850 M NaOH solution is required to neutralize 25.00 mL of H2SO4, what is the molarity of the acid solution?

Short Answer

Expert verified

The molarity of the acid solution is 0.097 M.

Step by step solution

01

Write the balanced molecular equation

The reaction of sodium hydroxide and sulfuric acid is shown below.

2NaOH(aq)+H2SO4(aq)→Na2SO4(aq)+2H2O(l)

02

Calculate the molarity of sulfuric acid

Given values are:

volumeofNaOHsolution=V(NaOH)=26.25mlConcentrationofNaOHsolution=S(NaOH)=0.1850 MvolumeofH2SO4=V(H2SO4)=25mlConcentrationofH2SO4solution=S(H2SO4)=?

From the balanced chemical reaction, it is seen that one mole of sulfuric acid neutralizes two moles of sodium hydroxide.

Therefore,

mole ratio=1 mol  H2SO42mol  NaOH=12

Using the following relation we can calculate the molarity of the acid solution.

V(H2SO4)×S(H2SO4)=V(NaOH)×S(NaOH)×1 mol (H2SO4) 2 mol(NaOH)⇒S(H2SO4)=V(NaOH)×S(NaOH)V(H2SO4)×12⇒S(H2SO4)=0.1850 M  ×26.25ml25ml×12⇒S(H2SO4)=0.097M

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