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Standard IX

Chemistry

pH of a Solution

Which one of ...

Question

Which one of following will have the largest pH?

Solution containing

1 \times 10^{- 2}

mol of

K_{2} S O_{4} L^{- 1}

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Pure water

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Solution containing

1.0 \times 10^{- 2}

mol of

H C l L^{- 1}

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Solution containing

1 \times 10^{- 2}

mol of

N H_{4} O H L^{- 1}

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Solution

The correct option is D Solution containing $1 \times 10^{- 2}$ mol of $N H_{4} O H L^{- 1}$
$p H = - l o g [H^{+}]$
$M / 100 H C l$
$p H = 2$
$10^{- 2}$ mol $N H_{4} O H$
$p H = 14 - p O H = 14 - 2 = 12$
Pure water and $K_{2} S O_{4}$ have $p H = 7$ .

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Similar questions

Q. Calculate the ratio of pH of a solution containing 1 mol of

C H_{3} C O O N a

+ 1 mol of

H C l

per litre to another solution containing 1 mol of

C H_{3} C O O N a

+ 1 mol of acetic acid per litre.

Q. Find the minimum concentration of

P b^{2 +}

ions that will initiate the precipitation of

P b S O_{4}

from a solution containing

0.005 m o l L^{- 1}

S O_{4}^{2 -}

ions?
(Given

K_{s p}

for

B a S O_{4} = 1.8 \times 10^{- 8}

)

Q. At

25^{\circ} C,

consider

1 L

of two solutions as following:
Solution A : Contains

1 m o l C H_{3} C O O N a + 0.5 m o l H C l

Solution B : Contains

1 m o l C H_{3} C O O N a + 0.5 m o l C H_{3} C O O H

K_{a} (C H_{3} C O O H) = 10^{- 5}

then ratio of [

H^{+}

(aq.)] of solution A to that of solution B is,
(Given:

l o g 1 = 0 a n d l o g 2 = 0.3, l o g (5 \times 10^{- 6} = - 5.3

)

Q. Calculate the ratio of pH of a solution containing 1 mol. of

C H_{3} C O O N a + 1 m o l

of HCl per litre and of other solution containing 1 mol of

C H_{3} C O O N a + 1 m o l

C H_{3} C O O H

per litre.

Q. Calculate the pH of the buffer containing

1.0 m o l L^{- 1}

of weak acid HA and

0.1 m o l L^{- 1}

of is sodium salt

N a A

assuming

K_{a}

to be

10^{- 6}

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Which one of following will have the largest pH?

The correct option is D Solution containing 1×10−2 mol of NH4OHL−1pH=−log[H+]M/100HClpH=210−2 mol NH4OHpH=14−pOH=14−2=12Pure water and K2SO4 have pH=7.

The correct option is D Solution containing $1 \times 10^{- 2}$ mol of $N H_{4} O H L^{- 1}$
$p H = - l o g [H^{+}]$
$M / 100 H C l$
$p H = 2$
$10^{- 2}$ mol $N H_{4} O H$
$p H = 14 - p O H = 14 - 2 = 12$
Pure water and $K_{2} S O_{4}$ have $p H = 7$ .