A 0.43g samle of KHP required 24.11cm of NaOH for neutralization. Calculate the molarity of NaOH

0.083 M

Explanation:

We'll begin by calculating the number of mole in 0.43 g of KHP (potassium hydrogen phthalate, C₈H₅O₄K). This is can be obtained as follow:

Mass of C₈H₅O₄K = 0.43 g

Molar mass of C₈H₅O₄K = (8×12) + (5×1) + (16× 4) + 39

= 96 + 5 + 64 + 39 = 204 g/mol

Mole of C₈H₅O₄K =?

Mole = mass / molar mass

Mole of C₈H₅O₄K = 0.43 / 204

Mole of C₈H₅O₄K = 0.002 mole

Next, we shall determine the number of mole of NaOH required to react with 0.43 g (i.e 0.002 mole) of KHP. This can be obtained as follow:

C₈H₅O₄K + NaOH → C₈H₄O₄KNa + H₂O

From the balanced equation above,

1 mole of KHP reacted with 1 mole of NaOH.

Therefore, 0.002 mole of KHP will also react with 0.002 mole of NaOH.

Next, we shall convert 24.11 cm³ to L. This can be obtained as follow:

1000 cm³ = 1 L

Therefore,

24.11 cm³ = 24.11 cm³ × 1 L / 1000 cm³

24.11 cm³ = 0.02411 L

Finally, we shall determine the molarity of NaOH. This can be obtained as follow:

Mole of NaOH = 0.00 2 mole

Volume = 0.02411 L

Molarity of NaOH =?

Molarity = mole /Volume

Molarity of NaOH = 0.002 / 0.02411

Molarity of NaOH = 0.083 M

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