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A 0.500g sample containing sodium dihydrogen phosphate is titrated with sodium hydroxide.if 23.06mL of 0.0985M sodium hydroxide is required for the titration, what is the percentage of NaHa2PO4 in the sample?
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+ Chemical reactions
+ Hydroxides
+ Sodium hydroxide
+ Titration
+ Sodium
+ Chemistry
Posted by
Larry Effler
A 0.500g sample containing sodium dihydrogen phosphate is titrated with sodium hydroxide.if 23.06mL of 0.0985M sodium hydroxide is required for the titration, what is the percentage of NaHa2PO4 in the sample?
The chemical formula for sodium dihydrogen phpsphate is "NaH₂PO₄" instead.
Molar mass of NaH₂PO₄ = (23 + 1×2 + 31 + 16×4) g/mol = 120 g/mol
Balanced equation for the reaction: NaH₂PO₄ + 2NaOH → Na₃PO₄ + 2H₂O Mole ratio NaH₂PO₄ : NaOH = 1 : 2
Moles of NaOH reacted = (0.0985 mol/L) × (23.06/1000 L) = 0.002271 mol Moles of NaH₂PO₄ reacted = (0.002271 mol) × (1/2) = 0.001136 mol Mass of NaH₂PO₄ in the sample = (0.001136 mol) × (120 g/mol) = 0.136 g Mass % of NaH₂PO₄ in the sample = (0.136/0.500) × 100% = 27.2% ==== OR:
(0.0985 mol NaOH / 1000 mL NaOH solution) × (23.06 mL NaOH solution) × (1 mol NaH₂PO₄ / 2 mol NaOH) × (120 g NaH₂PO₄ / 1 mol NaH₂PO₄) = 0.136 g NaH₂PO₄
The chemical formula for sodium dihydrogen phpsphate is "NaH₂PO₄" instead.
Molar mass of NaH₂PO₄ = (23 + 1×2 + 31 + 16×4) g/mol = 120 g/mol
Balanced equation for the reaction: NaH₂PO₄ + 2NaOH → Na₃PO₄ + 2H₂O Mole ratio NaH₂PO₄ : NaOH = 1 : 2
Moles of NaOH reacted = (0.0985 mol/L) × (23.06/1000 L) = 0.002271 mol Moles of NaH₂PO₄ reacted = (0.002271 mol) × (1/2) = 0.001136 mol Mass of NaH₂PO₄ in the sample = (0.001136 mol) × (120 g/mol) = 0.136 g Mass % of NaH₂PO₄ in the sample = (0.136/0.500) × 100% = 27.2% ==== OR:
(0.0985 mol NaOH / 1000 mL NaOH solution) × (23.06 mL NaOH solution) × (1 mol NaH₂PO₄ / 2 mol NaOH) × (120 g NaH₂PO₄ / 1 mol NaH₂PO₄) = 0.136 g NaH₂PO₄
The chemical formula for sodium dihydrogen phpsphate is "NaH₂PO₄" instead.
Molar mass of NaH₂PO₄ = (23 + 1×2 + 31 + 16×4) g/mol = 120 g/mol
Balanced equation for the reaction:
NaH₂PO₄ + 2NaOH → Na₃PO₄ + 2H₂O
Mole ratio NaH₂PO₄ : NaOH = 1 : 2
Moles of NaOH reacted = (0.0985 mol/L) × (23.06/1000 L) = 0.002271 mol
Moles of NaH₂PO₄ reacted = (0.002271 mol) × (1/2) = 0.001136 mol
Mass of NaH₂PO₄ in the sample = (0.001136 mol) × (120 g/mol) = 0.136 g
Mass % of NaH₂PO₄ in the sample = (0.136/0.500) × 100% = 27.2%
====
OR:
(0.0985 mol NaOH / 1000 mL NaOH solution) × (23.06 mL NaOH solution) × (1 mol NaH₂PO₄ / 2 mol NaOH) × (120 g NaH₂PO₄ / 1 mol NaH₂PO₄)
= 0.136 g NaH₂PO₄
(0.136 g NaH₂PO₄ / 0.500 g) × 100%
= 27.2% NaH₂PO₄
The chemical formula for sodium dihydrogen phpsphate is "NaH₂PO₄" instead.
Molar mass of NaH₂PO₄ = (23 + 1×2 + 31 + 16×4) g/mol = 120 g/mol
Balanced equation for the reaction:
NaH₂PO₄ + 2NaOH → Na₃PO₄ + 2H₂O
Mole ratio NaH₂PO₄ : NaOH = 1 : 2
Moles of NaOH reacted = (0.0985 mol/L) × (23.06/1000 L) = 0.002271 mol
Moles of NaH₂PO₄ reacted = (0.002271 mol) × (1/2) = 0.001136 mol
Mass of NaH₂PO₄ in the sample = (0.001136 mol) × (120 g/mol) = 0.136 g
Mass % of NaH₂PO₄ in the sample = (0.136/0.500) × 100% = 27.2%
====
OR:
(0.0985 mol NaOH / 1000 mL NaOH solution) × (23.06 mL NaOH solution) × (1 mol NaH₂PO₄ / 2 mol NaOH) × (120 g NaH₂PO₄ / 1 mol NaH₂PO₄)
= 0.136 g NaH₂PO₄
(0.136 g NaH₂PO₄ / 0.500 g) × 100%
= 27.2% NaH₂PO₄
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