Question

A student prepared a 20 mL sample of hydrochloric acid solution, The sample was diluted with distilled water. the buret was filled completely with the titrant, which is .1 M NaOH, until it reached the top mark. The sample was later titrated with a 0.10 M NaOH solution. The endpoint was reached after the addition of 15mL of titrant (NaOH). Calculate the concentration of HCL in molarity

also find these info:

Initial Volume reading of NaOH(mL)

Final Volume reading of NaOH(mL)

Volume of NaOH used(mL)

Molarity of NaOH(mol/L)

Moles of NaOH(mol)

Moles of HCL(mol)

Volume of HCL used (mL)

Molarity of HCL (mol/L)

Answer 1

Answer on Question #77423, Chemistry / Organic Chemistry

Question:

A student prepared a 20 mL sample of hydrochloric acid solution. The sample was diluted with distilled water. The buret was filled completely with the titrant, which is 0.1 M NaOH, until it reached the top mark. The sample was later titrated with a 0.10 M NaOH solution. The endpoint was reached after the addition of 15 mL of titrant (NaOH). Calculate the concentration of HCL in molarity.

Also find these info:

- Initial Volume reading of NaOH (mL)

- Final Volume reading of NaOH (mL)

- Volume of NaOH used (mL)

- Molarity of NaOH (mol/L)

- Moles of NaOH (mol)

- Moles of HCL (mol)

- Volume of HCL used (mL)

- Molarity of HCL (mol/L)

Solution:

- Volume of NaOH titrant: 15 mL = 0.015 L

- Amount of NaOH: 0.10 · 0.015 = 0.0015 mol

- Balanced equation:

$$\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}$$

- So the amount of HCl is same: 0.0015 mol

- Volume of a sample of HCl solution: 20 mL = 0.020 L

- Molarity of HCl:

$$\frac{0.0015}{0.020} = 0.075 \text{ mol/L}$$

Answer:

Concentration of HCl: 0.075 mol/L or 0.075 M

Additional info:

- Initial Volume reading of NaOH: 0 mL (from task)

- Final Volume reading of NaOH: 15 mL (from task)

- Volume of NaOH used: 15 mL (from task)

- Molarity of NaOH: 0.10 mol/L (from task)

- Moles of NaOH: 0.0015 mol (see solution)

- Moles of HCL: 0.0015 mol (see solution)

- Volume of HCL used: 20 mL (from task)

- Molarity of HCL: 0.075 mol/L (see solution)

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