If you have 0.15 equivalents of $H_{3} {P O}_{4}$ (full reaction or full neutralization occurs), 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 B 0.05
$E q u i v a l e n t w e i g h t = \frac{M o l a r m a s s}{n - f a c t o r}$
$N u m b e r o f g r a m e q u i v a l e n t s o f s o l u t e = \frac{G i v e n m a s s}{E q u i v a l e n t w e i g h t}$
$= \frac{G i v e n m a s s \times n - f a c t o r}{M o l a r m a s s}$
$∴ N u m b e r o f g r a m e q u i v a l e n t s o f s o l u t e = N u m b e r o f m o l e s \times n - f a c t o r$
$n - f a c t o r o f H_{3} P O_{4} = 3$ (Since $H$ is a tribasic acid)
Number of gram equivalents of solute $(H_{3} P O_{4}) = 0.15$ (given)
$M o l e s o f H_{3} P O_{4} = \frac{0.15}{3} = 0.05$

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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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