A solution of an unknown acid was prepared by dissolving 4.168 g of solid in eno

Question

A solution of an unknown acid was prepared by dissolving 4.168 g of solid in enough DI water to make 500.0 mL of solution. It takes 23.42 mL of 0.1678 M NaOH solution to neutralize 10.00 mL of the unknown acid solution.

A- Assuming that the acid is monoprotic. What is the concentration of the acid? what is the molar mass of the acid? ( hint: First calculate the total moles in 500 mL flask. The full 500 mL of solution was NOT titrated)

B- Now, assume that the acid is diporotic. What is the concentration of the acid? what is the molar mass of the acid?

Explanation / Answer

N1V1=N2V2

23.42*0.1678=N2*10.00

23.42*0.1678/10.00 =N2

N2== 0.3929876

if monoprotic M2=N2= 0.3929876 mol/l

total no of moles in solution= 0.3929876*0.5== 0.1964938 moles

so, molar mass= 4.168/0.1964938== 21.2118652 gm/mol

b.

if diprotic M2=N2/2=0.1964938 mol/l

total no of moles in solution= 0.1964938*0.5== 0.0982469 moles

so, molar mass= 4.168/0.0982469 = 42.4237304 gm/mol

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