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Standard IX

Chemistry

pH of a Solution

H 3 P O 4 is ...

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

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

$N a H_{2} P O_{4}$ basicity = 2

No of moles of $N a H_{2} P O_{4}$ = $\frac{12}{120}$

$E_{N a O H}$ = $E_{N a H_{2} P O_{4}}$

V $\times$ 1M = $\frac{12}{120}$ $\times$ 1000 $\times$ 2

V =200ml

Suggest Corrections

Similar questions

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 of

N a H_{2} P O_{4}

to convert it into

N a_{3} P O_{4}

H_{3} P O_{4}

is a tri basic acid and one of its salt is

N a H_{2} P O_{4}

. What volume in ml of

1

M 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}

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

P = 31

)

H_{3} P O_{4}

forms

N a H_{2} P O_{4}

N a_{2} H P O_{4}

and

N a_{3} P O_{4}

. Equivalent weight of acid at each stage of salt formation are

98

49

and

32.66

respectively.
If true enter 1, else enter 0.

Q. The equivalent weight of phosphoric acid

(H_{3} P O_{4})

in the reaction,

N a O H + H_{3} P O_{4} \to N a H_{2} P O_{4} + H_{2} O

is:

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H3PO4 is a tribasic acid and one of its salt is NaH2PO4. What volume of 1M NaOH solution should be added to 12 g ofNaH2PO4 to convert it intoNa3PO4?

The correct option is B 200 ml NaH2PO4 basicity = 2 No of moles of NaH2PO4 =12120 ENaOH=ENaH2PO4 V×1M =12120×1000×2 V =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 1M NaOH solution should be added to 12 g of $N a H_{2} P O_{4}$ to convert it into $N a_{3} P O_{4}$ ?

The correct option is B
200 ml

$N a H_{2} P O_{4}$ basicity = 2

No of moles of $N a H_{2} P O_{4}$ = $\frac{12}{120}$

$E_{N a O H}$ = $E_{N a H_{2} P O_{4}}$

V $\times$ 1M = $\frac{12}{120}$ $\times$ 1000 $\times$ 2

V =200ml