A solution of solution of NaOH was standardized by the procedure described in th

Question

A solution of solution of NaOH was standardized by the procedure described in this lab. An average of 42.12 ml of the base was required to titrate the 15.00 ml of 0.2878 M HCl standard to it's end point. the base was then used to determine the concentration of an unknown citric acid solution. citric acid has three acidic protons and ban be abbreviated H3Cit. Calculate the concentration of the unknown H3Cit solution if 10.00 ml of the solution required an average of 46.87 ml of the standardized base. (assume that all three protons in H3 Cit are titrated in this experiment)

I would really appreciate any help I can get!

Explanation / Answer

Volume of base,V1 = 42.12 mL

Volume of HCl, V2 = 15 mL

Molarity of HCl, M2 = 0.2878 M

Using M1*V1 = M2*V2

M1*42.12 = 0.2878*15

M1 = 0.1024 M

Now For titration of H3Cit -

Volume of base, V1 = 46.87

Volume of H3Cit, V2 = 10 mL

Let the concentration of unknown H3Cit solution be M2.

M1*V1 = M2*V2

0.1024*46.87 = M2*10

M2 = 0.4799 M

So, the concentration of the unknown H3Cit solution is 0.4799 M

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.