Standardizing a base solution by titration A chemistry student needs to standard

Question

Standardizing a base solution by titration A chemistry student needs to standardize a fresh solution of sodium hydroxide. She carefuly weighs out 42 img of oxalic acid (H,C,0,),a diprotic acid that can be purchased inexpensively in high punity, and dissoives t in 250. mlL of distlled water. The student then titrates the oxalic acid solution with her sodium hydroxide solution. When the titration reaches the equivalence point, the student finds she has used 11.1 mL of sodium hydroxide solution. Calculate the molarity of the student's sodium hydroxide solution significant digits . Be sure your answer has the correct number of 10P

Explanation / Answer

preparation of standard oxalic acid:

mass of oxalic acid=42 mg=42mg*(1g/1000mg)=0.042g

mol of oxalic acid=mass/molar mass=0.042g/90.03g/mol=4.665*10^-4 mol

molarity of oxalic acid=M1=mol of oxalic acid/volume of solution prepared=4.665*10^-4 mol/250ml=4.665*10^-4 mol/0.250L=0.001866 mol/L =0.001866M

volume of oxalic acid=V1=0.250L

molarity of NaOH=M2

Volume of NaOH used=V2=11.1ml*(1L/1000ml)=0.0111L

using equation,

mol of oxalic acid=M1V1

mol of NaOH=M2V2

1 mol of oxalic acid requires 2 mol of NaOH for neutralization,

2NaOH+H2C2O4---->Na2C2O4+2H2O

M1V1/M2V2=1/2

So,

M2=2*M1V1/V2=2*0.001866M*0.250L/0.0111L=2*0.0420 M=0.084M

M2=0.0840(molarity of NaOH)

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