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Reaction of Acids and Bases with Metals

H 3 PO 4 is a...

Question

$H_{3} {P O}_{4}$ is a tribasic acid and one of its salt is $N a H_{2} {P O}_{4}$ . What volume of 1M NaOH solution should be added to 12g $N a H_{2} {P O}_{4}$ to convert it into $N a_{3} {P O}_{4}$ ? (at.wt of P=31)

100 mL

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200 mL

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80 mL

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300 mL

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Solution

The correct option is B 200 mL
$N a H_{2} P O_{4}$ + $2 N a O H$ $\to N a_{3} P O_{4} + H_{2} O$
From the above reaction we can say that,
1 mol of $N a H_{2} P O_{4}$ reacts with 2 mol of NaOH.
Molar mass of $N a H_{2} P O_{4}$ = 120 g / mol
12g of $N a H_{2} P O_{4}$ contains = 0.1 mol
So the number of moles of NaOH required = 0.2 mol
Molarity of NaOH = 1M
1 mol = 1000 ml
0.2 mol = ?
So, the volume of 1M NaOH solution = 200 ml

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Similar questions

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

{N a}_{3} {P O}_{4}

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

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

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

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

p H = p K_{2} + {log}_{10} \frac{[H {P O}_{4}^{2 -}]}{[H_{2} {P O}_{4}^{-}]}

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

N a H_{2} {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}

50 m L

0.10 M

{N a}_{3} {P O}_{4} + 50 m L

0.10 M

N a H_{2} O_{4}

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

0.005 M

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

1 L

p H = 6.7

K_{1} = 7.1 \times 10^{- 3}

K_{2} = 6.3 \times 10^{- 8}

K_{3} = 4.5 \times 10^{- 13}

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

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