4. A 25.00 mL sample of HCl is titrated with the 0.315 M NaOH solution from the

Question

4. A 25.00 mL sample of HCl is titrated with the 0.315 M NaOH solution from the previous question. 22.65 mL of the NaOH solution was required to titrate the acid. Calculate the mo- larity ofthe HCl solution.…. von . . 5, A 1,08 g sample of an unknown monoprotic acid is titrated with 36.50 mL of 0.215 M NaOH. Calculate themolar mass oftheacid.:/ASgiad adHTSM 6. Sketch the molecular structure of acerysalicylic acid in aspirin and calculate its molar mass. Write the formula as CH,O 2. 7. A student dissolves a 0.726 g aspirin tablet in water and titrates the solution with 0.272 M NaOH. 12.05 mL of NaOH are required to reach the equivalence point. Calculate the per- centage of acetylsalicylic acid in the tablet.

Explanation / Answer

4)

Balanced chemical equation is:

NaOH + HCl ---> NaCl + H2O

Here:

M(NaOH)=0.315 M

V(NaOH)=22.65 mL

V(HCl)=25.0 mL

According to balanced reaction:

1*number of mol of NaOH =1*number of mol of HCl

1*M(NaOH)*V(NaOH) =1*M(HCl)*V(HCl)

1*0.315*22.65 = 1*M(HCl)*25.0

M(HCl) = 0.285 M

Answer: 0.285 M

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