Question

A NaOH solution is standardized using the monoprotic primary standard potassium hydrogen phthalate, KHP (204.22 g/mol.) If 0.6986 g of KHP requires 43.92 mL of NaOH, what is the NaOH concentration

Answer 1

Answer:

0.07789 M is the sodium hydroxide concentration.

Explanation:

Mass of potassium hydrogen phthalate = 0.6986 g

Molar mass of potassium hydrogen phthalate = 204.22 g/mol

Moles of  potassium hydrogen phthalate = $\frac{0.6986 g}{204.22 g/mol}=0.003421 mol$

$NaOH+KHP\rightarrow NaKP+H_2O$

According to reaction , 1 mole og KHp reactswith 1 mole of NaOH , then 0.003421 moles of KHp will react with :

$\frac{1}{1}\times 0.003421 mol=0.003421 mol$

Moles of NaOH = 0.003421 mole

Volume of NaOH solution = 43.92 ml = 0.04392 L ( 1 mL = 0.001L)

$Concentration=\frac{Moles}{Volume(L)}$

Concentration of NaOH :

$\frac{0.003421 mol}{0.04392 L}=0.07789 M$

0.07789 M is the sodium hydroxide concentration.