Question

A 0.2145g sample of KHP (FW = 204.22) was titrated with 27.12mL of NaOH. What is the molarity of the NaOH?

Answer 1

Task:

A 0.2145g sample of KHP (FW = 204.22) was titrated with 27.12mL of NaOH.

What is the molarity of the NaOH?

Solution:

The chemical equation for this reaction is

$$\mathrm{KHP} + \mathrm{NaOH} = \mathrm{KNaP} + \mathrm{H_2O}$$

The molarity is

$$C(\mathrm{M}) = n(\mathrm{mol}) / V(\mathrm{L})$$

n- amount of the substance (mol)

V – volume of solution (L)

According to the chemical equation the amount of KHP is equal to the amount of NaOH

$$(n(\mathrm{KHP}) = n(\mathrm{NaOH})$$ $$n(\mathrm{KHP}) = C(\mathrm{NaOH}) \cdot V(\mathrm{NaOH})$$

The amount of KHP is

$$n(\mathrm{KHP}) = m(\mathrm{KHP}) / \mathrm{FW}(\mathrm{KHP})$$

That's why

$$m(\mathrm{KHP}) / \mathrm{FW}(\mathrm{KHP}) = C(\mathrm{NaOH}) \cdot V(\mathrm{NaOH})$$

The molarity of NaOH is

$$C(\mathrm{NaOH}) = m(\mathrm{KHP}) / \left[\mathrm{FW}(\mathrm{KHP}) \cdot V(\mathrm{NaOH})\right] = 0.2145 / \left[204.22 \cdot 27.12 \cdot 10^{-3}\right] = 3.873 \cdot 10^{-2} \mathrm{M}$$

Answer: $C(\mathrm{NaOH}) = 3.873 \cdot 10^{-2} \mathrm{M}$