Home
Class 11
CHEMISTRY
25 mL of 0.4 M of a weak base and 75 mL ...

`25 mL` of `0.4 M` of a weak base and `75 mL` of `0.2` M of its salt forms a buffer solution. The dissocation constant of base is `2 xx 10^(-5)` the `(log 1.5 = 0.176)`

A

`4.876`

B

`9.476`

C

`4.524`

D

`9.124`

Text Solution

Verified by Experts

The correct Answer is:
D
`p^(H)=14-P_(kb)"log"([N_(s)xxV_(S)])/([N_(B)xxV_(B)])`
`=14-4.6"log"(75xx0.2)/(25xx0.4)=14-4.6-"log"(15)/(10)=9.12`
Promotional Banner

Topper's Solved these Questions

  • ACIDS & BASES

    NARAYNA|Exercise EXERCISE-II (Solubility Product)|5 Videos

  • ACIDS & BASES

    NARAYNA|Exercise EXERCISE-II (H.W.) (IONISATION OF ACIDS & BASES, DEGREE OF IONISATION & IONIC PRODUCT)|6 Videos

  • ACIDS & BASES

    NARAYNA|Exercise EXERCISE-II (Salt Hydrolysis)|3 Videos

  • 14TH GROUP ELEMENTS

    NARAYNA|Exercise Subjective Questions|6 Videos

  • ALKANES

    NARAYNA|Exercise Integer Answer Type|7 Videos

Similar Questions

Explore conceptually related problems

A solution containing 75 mL of 0.2M HCI and 25 mL of 0.2M NaOH. Calculate the pH of solution.

The pH of a buffer containing equal molar concentration of a weak base and its chloride (K_(b) weak base =2 xx 10^(-5) , log 2 =0.3 ) is

Calculate the pH of a solution obtained by diluting 1 mL of 0.10 M weak monoacidic base to 100 mL at constant temperature if Kb of the base is 1 xx 10^(-5) .

The dissociation constant of weak acid HA is 1 xx 10^(-5) . Its concentration is 0.1 M . pH of that solution is

50 mL of 0.2M ammonia solution is treated with 25 mL of 0.2 M HCl. If pK_(b) of ammonia solution is 4.75 the pH of the mixture will be :

50 ml of 0.05 M sodium hydroxide is mixed with 50 ml 0.1 of M acetic acid solution. What will be the pH resulting solution if K_(a) (CH_(3)COOH) = 2 xx 10^(-5)