A solution of sodium hydroxide (NaOH) was standardized against potassium hydroge

Question

A solution of sodium hydroxide (NaOH) was standardized against potassium hydrogen phthalate (KHP). A known mass of KHP was titrated with the NaOH solution until a light pink color appeared using phenolpthalein indicator. Using the volume of NaOH required to neutralize KHP and the number of moles of KHP titrated, the concentration of the NaOH solution was calculated. Molecular formula of Potassium hydrogen phthalate: HKC8H404 Mass of KHP used for standardization (g) 0.5181 Volume of NaOH required to neutralize KHP (mL) 13.4750 Calculate the concentration of NaOH solution in (mol/L)

Explanation / Answer

Hi dear Friend, you answer is as follows

Molar mass of KHP = 204.22 g/mol; moles = (m) mass / (M) Molar mass

Number of moles in 0.5181 grams of KHP = (m/M) = (0.5181g / 204.22 g/mol) mol = 0.00253696993 mol

KHC8H4O4 + NaOH H2O + NaKC8H4O4

Mole ratio = 1 KHP:1NaOH. hence both will reacts with equivalent moles i.e., moles KHP = moles NaOH

moles NaOH required = 0.00253696993 mol

Volume of NaOH = 13.475 mL = 0.013475 L

concentration of NaOH = moles/liter = 0.00253696993 mol / 0.013475 L = 0.18827235134 M

concentration of NaOH = 0.1883 M

Hope this helped you!

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