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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your Class 11 knowledge and support exam preparation. Choose a topic below to get started.
1. |
Assertion (A): Addition of `Ag^(o+)` ions to a mixture of aqueous `NaCI` and `NaBr` solution will first precipitate `AgBr` rather than `AgCI`. Reason (R) : `K_(sp) AgCI lt K_(sp) of AgBr`.A. If both (A) and (R) are correc, and (R) is the correct explanation of (A).B. If both (A) and (R) are correct but (R) is not the correct explanation of (A).C. If (A) is correct, but (R) is incorrect.D. If (A) is incorrect, but (R) is correct. |
Answer» Correct Answer - C<br>`(A)` is correct but `(R)` is false.<br> Correct `(R): K_(sp)` of `AgCI gt K_(sp)` of `AgBr` compounds with lower `K_(sp)` is precipitated first. Also, for precipiation `Q_(sp)` should be grater than `K_(sp)`. | |
2. |
`0.1M NaOH` is titrated with `0.1M HA` till the end point. `K_(a)` of HA is `5.6xx10^(-6)` and degree of hydrolysis is less compared to 1. Calculate pH of the resulting solution at the end point ? |
Answer» `{:(NaOH+,HArarr,NaA+,H_(2)O),(0.1,0.1,0,0),(0,0,0.1,0.1):}`<br> `[NaA] = (0.1)/(2)` because volume is doubled on mixing.<br> `NaA` is salt of weak acid `HA` and stron base `NaOH`.<br> For such salt solution:<br> `pH = 7 +(1)/(2) pK_(a) +(1)/(2)logC`<br> `= 7+(1)/(2)(-log 5.6 xx 10^(-5))+(1)/(2)log((0.1)/(2))`<br> `= 7+(1)/(2) (5-log 5.6) +(-0.65)`<br> `= 7 +(5.25)/(2) - 0.65 = 8.975` | |