Question

413 mg of dried KHP is dissolved in 41 mL of distilled water and titrated with potassium hydroxide (KOH). If it took 20.05 mL of KOH to reach the endpoint, determine the concentration of KOH. Show calculations.

Answer 1

Answer: The concentration of KOH is 0.1 M.

Explanation:

Given: Mass of KHP = 413 mg (1 mg = 0.001 g) = 0.413 g

Volume of KOH = 20.05 mL

Volume of solution = 41 mL

The molecular mass of KHP is 204.22 g/mol

Hence, moles of KHP are calculated as follows.

$No. of moles = \frac{mass}{molar mass}\\= \frac{0.413 g}{204.22 g/mol}\\= 0.002 mol$

As molarity is the number of moles of a substance present in a liter of solution.

So, molarity of the given solution is as follows.

$Molarity = \frac{no. of moles}{Volume (in L)}\\= \frac{0.002 M}{0.041 L}\\= 0.0493 M$

When the given solution is titrated with KOH then concentration of KOH is calculated as follows.

$M_{1}V_{1} = M_{2}V_{2}$

Substitute the values into above formula.

$M_{1}V_{1} = M_{2}V_{2}\\0.0493 M \times 41 mL = M_{2} \times 20.05 mL\\M_{2} = \frac{0.0493 M \times 41 mL}{20.05 mL}\\= 0.1 M$

Thus, we can conclude that the concentration of KOH is 0.1 M.