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Standard XII

Chemistry

Basic Buffer Action

How much Na...

Question

How much ${N a}_{2} H {P O}_{4}$ must be added to one litre of $0.005 M$ solution of $N a H_{2} {P O}_{4}$ in order to make $1 L$ of the solution of $p H = 6.7$ ?

$K_{1} = 7.1 \times 10^{- 3}$ ; $K_{2} = 6.3 \times 10^{- 8}$ ; $K_{3} = 4.5 \times 10^{- 13}$ for $H_{3} {P O}_{4}$

1.58 \times 10^{- 3}

mol

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3.1 \times 10^{- 3}

mol

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5.48 \times 10^{- 3}

mol

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None of these

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Solution

The correct option is A $1.58 \times 10^{- 3}$ mol
As we know, $p H = p K_{2} + log \frac{s a l t}{a c i d}$

Substituting, $6.7 = 7.2 + log \frac{y}{0.005}$

$y = 1.58 \times 10^{- 3}$ mole

So, $1.58 \times 10^{- 3}$ mol of $N a_{2} H P O_{4}$ must be added.

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Similar questions

Q. Calculate

[H^{+}]

[H_{2} {P O}_{4}]

[H P O_{4}^{2 -}]

and

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

in a

0.01 M

solution of

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

.
Take

K_{1} = 7.225 \times 10^{- 3}

K_{2} = 6.8 \times 10^{- 8}

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

Q. Calculate the

O H^{-}

concentration and the

H_{2} P O_{4}

concentration of a solution prepared by dissolving

0.1

mol of

N a_{3} P O_{4}

in sufficient water to make

1

L of solution.

K_{1} = 7.1 \times 10^{- 3}, K_{2} = 6.3 \times 10^{- 8}, K_{3} = 4.5 \times 10^{- 13}

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.

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_{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)

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How much Na2HPO4 must be added to one litre of 0.005M solution of NaH2PO4 in order to make 1L of the solution of pH=6.7? K1=7.1×10−3; K2=6.3×10−8; K3=4.5×10−13 for H3PO4

The correct option is A 1.58×10−3 molAs we know, pH=pK2+logsaltacidSubstituting, 6.7=7.2+logy0.005y=1.58×10−3 moleSo, 1.58×10−3 mol of Na2HPO4 must be added.

How much ${N a}_{2} H {P O}_{4}$ must be added to one litre of $0.005 M$ solution of $N a H_{2} {P O}_{4}$ in order to make $1 L$ of the solution of $p H = 6.7$ ?

$K_{1} = 7.1 \times 10^{- 3}$ ; $K_{2} = 6.3 \times 10^{- 8}$ ; $K_{3} = 4.5 \times 10^{- 13}$ for $H_{3} {P O}_{4}$

The correct option is A $1.58 \times 10^{- 3}$ mol
As we know, $p H = p K_{2} + log \frac{s a l t}{a c i d}$

Substituting, $6.7 = 7.2 + log \frac{y}{0.005}$

$y = 1.58 \times 10^{- 3}$ mole

So, $1.58 \times 10^{- 3}$ mol of $N a_{2} H P O_{4}$ must be added.