Potassium hydrogen phthalate, KHP, is a monoprotic acid often used to standardize NaOH solutions. If 0.212 g of KHP are dissolved into 50.00 mL of water and titrated with 35.00 mL of NaOH, what is the molarity of NaOH. Molar mass of KHP

Answers

Answer 1

The molarity of the NaOH solution is 0.03 M

We'll begin by calculating the mole of the KHP

Mass = 0.212 g Molar mass = 204.22 g/mol Mole of KHP =?

Mole = mass /molar mass

Mole of KHP = 0.212 / 204.22

Mole of KHP = 0.001 mole

Next, we shall determine the molarity of the KHP solution

Mole of KHP = 0.001 mole Volume = 50 mL = 50/1000 = 0.05 L Molarity of KHP =?

Molarity = mole / Volume

Molarity of KHP = 0.001 / 0.05

Molarity of KHP = 0.02 M

Finally , we shall determine the molarity of the NaOH solution

KHP + NaOH —> NaPK + H₂O

From the balanced equation above,

The mole ratio of the acid, KHP (nA) = 1The mole ratio of base, NaOH (nB) = 1

From the question given above, the following data were obtained:

Volume of acid, KHP (Va) = 50 mL Molarity of acid, KHP (Ma) = 0.02 M.Volume of base, NaOH (Vb) = 35 mL Molarity of base, NaOH (Mb) =?

MaVa / MbVb = nA / nB

(0.02 × 50) / (Mb × 35) = 1

1 / (Mb × 35) = 1

Cross multiply

Mb × 35 = 1

Divide both side by 35

Mb = 1 / 35

Mb = 0.03 M

Thus, the molarity of the NaOH solution is 0.03 M

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Potassium Hydrogen Phthalate, KHP, Is A Monoprotic Acid Often Used To Standardize NaOH Solutions. If

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We are given that the molar mass of boron triflouride BF₃ is 67.81 g/mol. We want to determine the number of molecules of BF₃ in 2 g of BF₃.

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The number of grams of ammonium nitrate (NH₄NO₃) it would take for all the barium hydroxide to react is 18.7g

First, we will write the balanced chemical equation for the reaction

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This means, 1 mole of barium hydroxide is required to react with 2 moles of ammonium nitrate

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[tex]Number\ of\ moles = \frac{Mass}{Molar\ mass}[/tex]

Molar mass of Ba(OH)₂ = 171.34 g/mol

∴ Number of moles of Ba(OH)₂ present =[tex]\frac{20}{171.34}[/tex]

Number of moles of Ba(OH)₂ present = 0.116727 mole

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Since 1 mole of barium hydroxide is required to react with 2 moles of ammonium nitrate

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0.116727 mole of barium hydroxide will react with 2 × 0.116727 mole of ammonium nitrate

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