What is the different between a primary and a secondary standary? Give an exampl

Question

What is the different between a primary and a secondary standary? Give an example of each. Why must air bubbles be expelled from the buret tip? 0.3045 g of pure KHP was weighed out and titrated to an end (mint with 15.12 mL of a NaOH solution that was approximately 0.1 M. What is the exactly concentration of the NaOH titrant? 0.5366 g of an KHP sample of unknown purity was massed. Die sample was dissolved in approximately 100 mL of distilled, degassed water and indicator was added. The end point was reached after 21.35 mL of 0.09854 M NaOH solution was titrated into the solution. What is the percentage of KHP in the original sample? What is the molarity of a NaOH solution if 32.47 mL is required (o titrate 0.6013 g of potassium hydrogen phthalate (KHCJH4O4)?

Explanation / Answer

1. Primary standard is the solution whose concentration doesnot change with respect to time. Features of a primary standard include: hIigh purity, stability, Low hygroscopicity, High equivalent weight , Non-toxicity, Ready and cheap availability.

Eg: standard solution of oxalic acid, KHP, or Potassium bromate (KBrO3).

Secondary standard is the solution whose concentration changes with respect to time.

Eg: Solutions of NaOH, HCl, or H2SO4

2. The presence of air bubles in the burette tip directly effects on the accurate measurment of titrant volume, thus, burette tip should be free from air bubles before starting the titration.

3. Weight of KHP= 0.3043 g

Molecular weight of KHP= 204.22 g/mol

WKT

Molarity of NaOH= (weight of KHP/ Molecular weight of KHP) X (1000 mL / Volume of NaOH)

Molarity of NaOH = (0.3043 g/ 204.22 g/mol) X (1000 mL / 15.12 mL)

Molarity of NaOH = 0.09854 M

4.

WKT, (MV) NaOH = (MV) KHP

0.09854 M x 21.35 mL = M KHP x 100 mL

or M KHP= 0.0210 M

WKT

M KHP = (weight of KHP/ Molecular weight of KHP) x (1000 mL / volume of KHP)

Weight of KHP = (M KHP x Molecualr weight of KHP x volume of KHP) / 1000 mL

Weight of KHP = 0.0210 M x 204.22 g/mol x 100 mL / 1000 mL

Weight of KHP = 0.4288 g is present in 100 mL

Percentage purity = (weight of KHP/ Weight of KHP unknown purity) x 100

Percentage purity = (0.4288 g / 0.5366 g) x 100

Percentage purity of KHP = 79.92 %

5.

WKT

Molarity of NaOH= (weight of KHP/ Molecular weight of KHP) X (1000 mL / Volume of NaOH)

Molarity of NaOH = (0.6013 g / 204.22 g/mol) X (1000 mL / 32.47 mL)

Molarity of NaOH = 0.0906 M is required.

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

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