A $31.27 \cdot m L$ $31.27mL$ volume of sodium hydroxide at $0.167 \cdot m o l \cdot L^{- 1}$ $0.167molL^-1$ concentration was used to neutralize a $22.5 \cdot m L$ $22.5mL$ volume of AQUEOUS $H_{3} P O_{4}$ $H_3PO_4$ . What is $[H_{3} P O_{4}] (a q)$ $[H_3PO_4](aq)$ ?

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A $31.27 \cdot m L$ $31.27mL$ volume of sodium hydroxide at $0.167 \cdot m o l \cdot L^{- 1}$ $0.167molL^-1$ concentration was used to neutralize a $22.5 \cdot m L$ $22.5mL$ volume of AQUEOUS $H_{3} P O_{4}$ $H_3PO_4$ . What is $[H_{3} P O_{4}] (a q)$ $[H_3PO_4](aq)$ ?

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Answer 1 · 2016-08-26T15:35:04

$H_{3} P O_{4} (a q) + 2 N a O H (a q) \to N a_{2} H P O_{4} (a q)$ $H_3PO_4(aq) + 2NaOH(aq) rarr Na_2HPO_4(aq)$

Explanation:

Note the stoichiometry. You reach a stoichiometric endpoint at biphosphate, $H P O_{4}^{2 -}$ $HPO_4^(2-)$ , NOT $P O_{4}^{3 -}$ $PO_4^(3-)$ .

$Moles of sodium hydroxide$ $"Moles of sodium hydroxide"$ $=$ $=$ $31.27 \times 10^{- 3} L \times 0.167 \cdot m o l \cdot L^{- 1}$ $31.27xx10^-3Lxx0.167*mol*L^-1$ $=$ $=$ $5.22 \times 10^{- 3} \cdot m o l$ $5.22xx10^-3*mol$ .

There were thus $\frac{5.22 \times 10^{- 3} \cdot m o l}{2}$ $(5.22xx10^-3*mol)/2$ with respect to $H_{3} P O_{4}$ $H_3PO_4$ in the volume of $22.5 \cdot m L$ $22.5*mL$ .

${Molarity}_{phosphoric acid}$ $"Molarity"_"phosphoric acid"$ $=$ $=$ $\frac{5.22 \times 10^{- 3} \cdot m o l}{2} \times \frac{1}{22.5 \times 10^{- 3} L}$ $(5.22xx10^-3*mol)/2xx1/(22.5xx10^-3L)$ $≅$ $~=$ $0.1 \cdot m o l \cdot L^{- 1}$ $0.1*mol*L^-1$ with respect to phosphoric acid.

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A $31.27 \cdot m L$ $31.27mL$ volume of sodium hydroxide at $0.167 \cdot m o l \cdot L^{- 1}$ $0.167molL^-1$ concentration was used to neutralize a $22.5 \cdot m L$ $22.5mL$ volume of AQUEOUS $H_{3} P O_{4}$ $H_3PO_4$ . What is $[H_{3} P O_{4}] (a q)$ $[H_3PO_4](aq)$ ?

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A 31.27⋅mL31.27*mL volume of sodium hydroxide at 0.167⋅mol⋅L−10.167*mol*L^-1 concentration was used to neutralize a 22.5⋅mL22.5*mL volume of AQUEOUS H3PO4H_3PO_4. What is [H3PO4](aq)[H_3PO_4](aq)?

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A $31.27 \cdot m L$ $31.27mL$ volume of sodium hydroxide at $0.167 \cdot m o l \cdot L^{- 1}$ $0.167molL^-1$ concentration was used to neutralize a $22.5 \cdot m L$ $22.5mL$ volume of AQUEOUS $H_{3} P O_{4}$ $H_3PO_4$ . What is $[H_{3} P O_{4}] (a q)$ $[H_3PO_4](aq)$ ?