A damaged bottle label has made life difficult for a friendly chemist. The bottl

Question

A damaged bottle label has made life difficult for a friendly chemist. The bottle contains 325 mL of an NaOH solution but the concentration is no longer readable.
The chemist removes 5.00 mL of the NaOH solution using a volumetric pipette. Then she adds water and titrates the resulting solution with hydrochloric acid of known concentration. From this she determines that there were 0.0113 moles of NaOH in the 5.00 mL sample.
a. How many moles of NaOH are there in the original 325 mL solution?
b. What is the concentration of the original NaOH solution?
A damaged bottle label has made life difficult for a friendly chemist. The bottle contains 325 mL of an NaOH solution but the concentration is no longer readable.
The chemist removes 5.00 mL of the NaOH solution using a volumetric pipette. Then she adds water and titrates the resulting solution with hydrochloric acid of known concentration. From this she determines that there were 0.0113 moles of NaOH in the 5.00 mL sample.
a. How many moles of NaOH are there in the original 325 mL solution?
b. What is the concentration of the original NaOH solution?
A damaged bottle label has made life difficult for a friendly chemist. The bottle contains 325 mL of an NaOH solution but the concentration is no longer readable.
The chemist removes 5.00 mL of the NaOH solution using a volumetric pipette. Then she adds water and titrates the resulting solution with hydrochloric acid of known concentration. From this she determines that there were 0.0113 moles of NaOH in the 5.00 mL sample.
a. How many moles of NaOH are there in the original 325 mL solution?
b. What is the concentration of the original NaOH solution?
A damaged bottle label has made life difficult for a friendly chemist. The bottle contains 325 mL of an NaOH solution but the concentration is no longer readable.
The chemist removes 5.00 mL of the NaOH solution using a volumetric pipette. Then she adds water and titrates the resulting solution with hydrochloric acid of known concentration. From this she determines that there were 0.0113 moles of NaOH in the 5.00 mL sample.
a. How many moles of NaOH are there in the original 325 mL solution?
b. What is the concentration of the original NaOH solution?
A damaged bottle label has made life difficult for a friendly chemist. The bottle contains 325 mL of an NaOH solution but the concentration is no longer readable.
The chemist removes 5.00 mL of the NaOH solution using a volumetric pipette. Then she adds water and titrates the resulting solution with hydrochloric acid of known concentration. From this she determines that there were 0.0113 moles of NaOH in the 5.00 mL sample.
a. How many moles of NaOH are there in the original 325 mL solution?
b. What is the concentration of the original NaOH solution?

Explanation / Answer

a) There are 0.0113 mole NaOH in 5.00 mL of the sample. 5.00 mL of the NaOH sample was removed from 325 mL of the original sample. Hence, the dilution factor = (volume of original sample)/(volume of sample taken) = (325 mL)/(5.00 mL) = 65.

Mole(s) of NaOH in the original 325 mL sample solution = (0.0113 mole)*(dilution factor) = (0.0113 mole)*(65) = 0.7345 mole (ans).

b) Concentration of the original NaOH solution = (moles of NaOH)/(volume of the solution in L) = (0.7345 mole)/[(325 mL)*(1 l/1000 mL)] = 2.26 mol/L = 2.26 M (ans).

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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