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pH of Salts

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 $1 M N a O H$ solution should be added to $12 g {N a H}_{2} {P O}_{4}$ to convert it into ${N a}_{3} {P O}_{4}$ ? ( $a t . w t$ of $P = 31$ )

100 m l

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200 m l

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80 m l

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300 m l

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Solution

The correct option is A $200 m l$
$N a H_{2} {P O}_{4} + N a O H \to {N a}_{2} {H P O}_{4} + H_{2} O - (i)$
Moles $= \frac{12}{120}$ $\frac{12}{120}$ $\frac{12}{120}$
${N a}_{2} {H P O}_{4} + N a O H \to {N a}_{3} {P O}_{4} + H_{2} O - (i i)$
Moles $= \frac{12}{120}$ $\frac{12}{120}$ $\frac{12}{120}$
Moles of $N a O H$ required $= \frac{12}{120} + \frac{12}{120}$
$= 0.2$ moles
Now, $M u l a r i t y = \frac{M o l e s}{V o l . o f s o l i n l t r s}$
$∴$ Volume of $N a O H$ solution $= \frac{0.2}{1}$
$= 0.2$ litres
$= 200$ ml

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Similar questions

Q. To convert 12 g of

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

complete to

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

, volume of 1 M NaOH solution required is:

$H_{3} P O_{4}$ 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 12 g of $N a H_{2} P O_{4}$ to convert it into $N a_{3} P O_{4}$ ?

Q. Orthophosphoric acid

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

is a weak tribasic mineral acid,

H_{3} {P O}_{4} ⇌ H^{+} + H_{2} {P O}_{4}^{-}; K_{a_{1}} = 10^{- 3}

H_{2} {P O}_{4}^{-} ⇌ H^{+} + H {P O}_{4}^{- 2}; K_{a_{2}} = 10^{- 7}

H P O_{4}^{- 2} ⇌ H^{+} + {P O}_{4}^{- 3}; K_{a_{3}} = 10^{- 13}

In a solution of

0.1 M

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

, concentration of

{P O}_{4}^{3 -}

is:

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

Hence, pH of

0.01 M

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

is:

Q. In a buffer solution consisting

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

and

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

(1)

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

is acid and

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

is salt
(2) The pH of solution can be calculated using the relation:

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

(3) The

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

is acid and

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

is salt
(4) The pH cannot be calculated.

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H3PO4 is a tribasic acid and one of its salt is NaH2PO4 What volume of 1MNaOH solution should be added to 12g NaH2PO4 to convert it into Na3PO4? (at.wt of P=31)

The correct option is A 200 mlNaH2PO4+NaOH→Na2HPO4+H2O−(i)Moles=12120 12120 12120Na2HPO4+NaOH→Na3PO4+H2O−(ii)Moles=12120 12120 12120Moles of NaOH required =12120+12120 =0.2 molesNow, Mularity=MolesVol.ofsolinltrs∴ Volume of NaOH solution =0.21 =0.2 litres =200ml

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