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How many milliliters of a 0.1500 – M solution of KOH will be required to titrate 40.00 ML’s of 0.0656 – M solution of H3 PO four

How many milliliters of a 0.1500 – M solution of KOH will be required to titrate 40.00 ML-example-1
User KevInSol
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1 Answer

3 votes

Step 1

The reaction:

H3PO4 (aq) + 2 KOH (aq) => K2HPO4 (aq) + 2 H2O (l)

The equation must be always completed and balanced.

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Step 2

Information provided:

H3PO4) 40.00 mL of 0.0656 M

40.00 mL x (1 L/1000 mL) = 0.0400 L

KOH) 0.1500 M

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Step 3

The number of moles of H3PO4:

Molarity (M or mol/L) = moles/volume (L)

Molarity x volume (L) = moles

0.0656 mol/L x 0.0400 L = 2.62x10^-3 moles H3PO4

The number of moles of H3PO4 = 2.62x10^-3 moles

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Step 4

The number of moles of KOH: by stoichiometry,

H3PO4 (aq) + 2 KOH (aq) => K2HPO4 (aq) + 2 H2O (l)

1 mole of H3PO4 ------------ 2 moles KOH

2.62x10^-3 moles H3PO4 ------------ X

X = 2.62x10^-3 moles H3PO4 x 2 moles KOH/1 mole of H3PO4

X = 5.24x10^-3 moles KOH

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Step 5

The volume of KOH:

Molarity = moles/volume (L)

Volume (L) = moles/molarity

Volume (L) = 5.24x10^-3 moles/0.1500 mol/L

Volume = 0.0349 L x (1000 mL/1 L) = 34.9 mL

Answer: 34.9 mL

User Nazar Kalytiuk
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