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The P^(H) of HCI is 1. The amount of NaO...

The `P^(H)` of HCI is 1. The amount of NaOH to be added to 100 ml of such a HCI solution to get `p^(H)` of 7 is

A

4g

B

`0.4 ` g

C

4 mg

D

`0.4 mg

Text Solution

Verified by Experts

The correct Answer is:
B
To get pH = 7
no. of eq. . Of base = no. of e eq. of acid
pH of HCl = 1 ` [H^(+)= 0.01 N , [HCl] = 0.1 N`
From (1) , [NaOH] = 0.1 N
` 0.1 = (W)/( 40 ) xx ( 1000)/(100) therefore W = 0.4 gm `
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AAKASH SERIES-IONIC EQUILIBRIUM -LECTURE SHEET (EXERCISE -II LEVEL -II (ADVANCED) (Straight Objective Type Questions))
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  2. The P^(H) of 40 ml of 0.02 M HCl will not be changed by adding

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  3. The P^(H) of HCI is 1. The amount of NaOH to be added to 100 ml of suc...

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  4. How many grams of NaOH can neutralise 100 ml of H2SO4 solution with ...

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  5. 20 ml of 0.4 M H(2),SO(4) , and 80 ml of 0.2 M NaOH are mixed. Then th...

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  6. x xx 10^(-2) gm of NaOH should be used up to prepare 200ml of a soluti...

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  7. What is the concentration of CN^(-) ions in a solution with 0.1 M HC...

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  8. P^(H) of a solution of the mixture of 0.1 HCl and 0.1N CH3 COOH is (...

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  9. Given HF + H2O overset( Ka) hArr H3 O^(+) +F ^(-) , F^(-) +H2O over...

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  10. Liquid ammonia ionises to a slight extent. Its self ionisation constan...

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  11. The P^(H )of a solution is 6. Its [H(3),O^(4)] is decreased by 1000 ti...

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  12. The P^(H) of a solution is 11. lt is diluted by 1000 times. Then the P...

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  13. The P^(H) of 40 ml of 0.02 M HCl will not be changed by adding

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  14. The P^(H) of HCI is 1. The amount of NaOH to be added to 100 ml of suc...

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  15. How many grams of NaOH can neutralise 100 ml of H2SO4 solution with ...

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  16. 20 ml of 0.4 M H(2),SO(4) , and 80 ml of 0.2 M NaOH are mixed. Then th...

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  17. Among the following (a) On dilution the P^(H) of an acid increases...

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  18. What is the concentration of CN^(-) ions in a solution with 0.1 M HC...

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  19. P^(H) of a solution of the mixture of 0.1 HCl and 0.1N CH3 COOH is (...

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  20. Given HF + H2O overset( Ka) hArr H3 O^(+) +F ^(-) , F^(-) +H2O over...

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