If a student have 0.15 equivalents of $H_{3} {P O}_{4}$ (full reaction or full neutralization occurs), then how many moles of $H_{3} {P O}_{4}$ do you have?

0.05

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0.15

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4.9

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9.8

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Solution

The correct option is E 0.05
$H_{3} P O_{4}$ is tribasic acid, means it can release 3 $H^{+}$ from the $- O H$ groups attached to the $P$ .
$H_{3} P O_{4} \to 3 H^{+} + P O_{4}^{3 -}$
Let moles of $H_{3} P O_{4}$ be $n$ .
$0.15 = 3 \times n$
$n = \frac{0.15}{3} = 0.05 m o l e s$

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Similar questions

H_{3} {P O}_{4}

H_{3} {P O}_{4}

H_{2} {P O}_{4}^{-}

H_{3} {P O}_{4} + H_{2} O \to H_{3} O^{+} + H_{2} {P O}_{4}^{-}

H_{2} {P O}_{4}^{-} + H_{2} O \to H {P O}_{4}^{2 -} + H_{3} O^{+}

H_{2} {P O}_{4}^{-} + {O H}^{-} \to H_{3} {P O}_{4} + O^{2 -}

H_{3} {P O}_{4}

H_{2} O + H_{3} {P O}_{4} ⇌ H_{3} O^{+} + H_{2} {P O}_{4}^{-}; p K_{1} = 2.15

H_{2} O + H_{2} {P O}_{4}^{-} ⇌ H_{3} O^{+} + H_{2} {P O}_{4}^{2 -}; p K_{2} = 7.20

0.01 M

N a H_{2} {P O}_{4}

1 g

H_{3} {P O}_{4}

3 g

N a O H

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