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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 601. |
Which of the following is not Lewis acidA. `AlCl_(3).6H_(2)O`B. `AlCl_(3)`C. `SnCl_(4)`D. `FeCl_(3)` |
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Answer» Correct Answer - A `AlCl_(3)` acts as a lewis acid as central atom Al has incomplete octet. `SnCl_(4)` and `FeCl_(3)` act as lewis acid because Sn and Fe both have empty d - orbitals. `AlCl_(3)` is `sp^(2)` hybridised, i.e., it has vacant one 3p orbital and five 3d orbitals. But in `AlCl_(3).6H_(2)O` one vacant 3p orbital and five vacant 3d orbitals are filled by the electron pair of oxygen. So `AlCl_(3).6H_(2)O` cannot act as a lewis acid. |
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| 602. |
In a solution of `0.04M FeCl_(2)` and `0.01 M FeCl_(3)`, how large may be its pH of without being precipitation of either `Fe(OH)_(2)` or `Fe(OH)_(3)` ? [Given `K_(sp) Fe(OH)_(2) = 16 xx 10^(-6)` and `K_(sp) Fe(OH)_(3) = 8 xx 10^(-26)`]A. `5.7`B. `6.3`C. `8.3`D. `10.7` |
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Answer» Correct Answer - B For the precipitation of `Fe(OH)_(2)` `K_(sp) = [0.04][OH^(-)]^(2) = 16 xx 10^(-6)` `[OH^(-)] = 2 xx 10^(-2)` and for `Fe[OH]_(3)` `K_(sp) = 0.02 xx [OH^(-)]^(2) = 16 xx 10^(-6)` `[OH^(-)] = 2 xx 10^(-2)` For `Fe(OH)_(3)` the `[OH^(-)]` ions are required less `pOH = 8 - "log" 2` `pH = 6 + "log" 2 = 6.3` |
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| 603. |
`CH_(3)COOH` is weaker acid than `H_(2)SO_(4)`. It is due toA. More ionizationB. Less ionizationC. Covalent bondD. Electrovalent bond |
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Answer» Correct Answer - B The `CH_(3)COOH` is weaker acid than `H_(2)SO_(4)`. The reason is that `CH_(3)COOH` is weakly ionised in comparison with `H_(2)SO_(4)`. |
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| 604. |
In the reaction `NH_(3) + H_(2)O hArr NH_(4)^(+) + barOH`, the conjugate acid-base pair isA. `NH_(3)` and `H_(2)O`B. `NH_(3)` and `OH^(-)`C. `H_(2)O` and `NH_(4^(+))`D. `NH_(4^(+)) ` and `NH_(3)` |
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Answer» Correct Answer - D Conjugate acid-base pair is differ by only one proton |
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| 605. |
A solution of `FeCl_(3)` in water acts as acidic due toA. Hydrolysis of `Fe^(3+)`B. Acidic impuritiesC. DissociationD. Ionisation |
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Answer» Correct Answer - A `FeCl_(3) + 3H_(2)O hArr Fe(OH)_(3) + 3HCl`. Strong acid and weak base. |
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| 606. |
A solution of sodium bicarbonate in water turnsA. Phenolphthalein pinkB. Methyl orange yellowC. Methyl orange redD. Blue litmus red |
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Answer» Correct Answer - C `NaHCO_(3)` in water is alkaline in nature due to hydrolysis of `HCO_(3)^(-)` ion. `NaCO_(3) hArr Na^(+) + HCO_(3)^(-)`. |
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| 607. |
A 0.1 N solution of an acid at room temperature has a degree of ionisation 0.1. The concentration of `OH^(-)` would beA. `10^(-12) M`B. `10^(-11) M`C. `10^(-9) M`D. `10^(-2) M` |
| Answer» Correct Answer - A | |
| 608. |
Which of the following solutions has the highest pH?A. `CH_(3)COOK`B. `Na_(2)CO_(3)`C. `NH_(4)Cl`D. `NaNO_(3)` |
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Answer» Correct Answer - B `NH_(4)Cl` is acidic , its `pH lt 7 . NaNO_(3)` solution is neutral its `pH =7CH_(3)COOK` and `Na_(2)CO_(3)` solutions are basic their `pH lt7`. `Na_(2)CO_(3)` solution is more basic, its `Ph gt pH` of `CH_(3)COOK` solution. |
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| 609. |
If the pH of a solution of an alkali metal hydroxide is 13.6, the concentration of hydroxide isA. Between 0.1 M and 1MB. More than 1 MC. Less than 0.001 MD. Between 0.01 M and 1 M |
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Answer» Correct Answer - A `pH = 13.6` pOh = 14 - 13.6 = 0.4 `[OH^(-)]` = Antilog (0.4) = 0.3979. So the value of `[OH^(-)]` between 0.1 M and 1 M. |
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| 610. |
Arrange `NH_(4)^(+), H_(2)O, H_(3)O^(+), HF` and `OH^(-)` in increasing order of acidic natureA. `H_(3)O^(+) lt NH_(4)^(+) lt HF lt OH^(-) lt H_(2)O`B. `NH_(4)^(+) lt HF lt H_(3)O^(+) lt H_(2)O lt OH^(-)`C. `OH^(-) lt H_(2)O lt NH_(4)^(+) lt HF lt H_(3)O^(+)`D. `H_(3)O^(+) gt HF gt H_(2)O gt NH_(4)^(+) gt OH^(-)` |
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Answer» Correct Answer - C `H_(3)O^(+) gt HF gt NH_(4)^(+) + H_(2)O gt OH^(-)`. Acidic nature in decreasing order. |
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| 611. |
Arrange `NH_(4)^(+), H_(2)O, H_(3)O^(+), HF` and `OH^(-)` in increasing order of acidic nature:A. `OH^(-)ltH_(2)OltNH_(4)^(+)ltHFltH_(3)O^(+)`B. `H_(3)O^(+)ltHFltH_(2)OltNH_(4)^(+)ltOH^(-)`C. `NH_(4)^(+)ltHFltH_(3)O^(+)ltH_(2)OltOH^(-)`D. `H_(3)O^(+)ltNH_(4)^(+)ltHFltOH^(-)ltH_(2)O` |
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Answer» Correct Answer - A `OH^(-)` is base and `H_(2)O` is neutral. |
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| 612. |
In aqueous solution the ionization constant for carbonic acid are `K_(1) = 4.2 xx 10^(-7)` and `K_(2) = 4.8 xx 10^(-11)` Select the correct statement for a saturated 0.034 M solution of the carbonic acidA. The concentration of `H^(+)` is double that of `CO_(3)^(2-)`B. The concentration of `CO_(3)^(2-)` is 0.034 MC. The concentration of `CO_(3)^(2-)` is greater than that of `HCO_(3)^(-)`D. The concentration of `H^(+)` and `HCO_(3)^(-)` are approximately equal |
| Answer» Correct Answer - D | |
| 613. |
Which of the following has highest value of `K_(sp)` ?A. `Mg(OH)_(2)`B. `Ca(OH)_(2)`C. `Ba(OH)_(2)`D. `Be(OH)_(2)` |
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Answer» Correct Answer - D from `Be(OH)_(2)` to `Ba(OH)_(2)` solubility product increases `K_(SP) = [Pb^(+2)][I^(-)]` |
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| 614. |
The pH of neutral water is 6.5 , then the temperature of water isA. `25^(@)C`B. more than `25^(@)C`C. can be more or less than `25^(@)C`D. cannot be predicted |
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Answer» Correct Answer - B pH less than 7 means `[H^(+)]` greater than `10^(-7)`. Which is so if the temperature is more than `25^(@)C`. |
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| 615. |
Which solutions has the highest pH valueA. 1M KOHB. `1M H_(2)SO_(4)`C. Chlorine waterD. Water containing carbon dioxide |
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Answer» Correct Answer - A 1 MKOH show highest pH value because it is a strong base. |
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| 616. |
Which of the following mixtures can act as a buffer?A. `NaOH + HCOONa(1:1 "molar ratio")`B. `HCOOH + NaOH(2:1 "molar ratio")`C. `NH_(4)Cl + NaOH(2:1 "molar ratio")`D. `HCOOH + NaOH(1:1 "molar ratio")` |
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Answer» Correct Answer - B::C `{:("For C option",NH_(4)Cl,+,NaOH,rarr,NH_(4)OH+NaCl),("Initial",2,,1,,0),(R&F:,1,,1,,1),("Left:",1,,0,,1),("Hence basic buffer:",,,,,):}` |
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| 617. |
Concentration of CN in 0.1 M HCN is `[K_(a) = 4 xx 10^(-10)]`A. `2.5 xx 10^(-6)M`B. `4.5 xx 10^(-6)M`C. `6.3 xx 10^(-6)M`D. `9.2 xx 10^(-6)M` |
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Answer» Correct Answer - C Concentration of `CN^(-) = sqrt(K_(a)C) = 6.3 xx 10^(-6)`. |
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| 618. |
As the temperature increases, the pH of a KOH solutionA. Will decreasesB. Will increasesC. Remains constantD. Depends upon concentration of KOH solution |
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Answer» Correct Answer - A pH will decrease because `[OH^(-)]` increased due to this pOH is decreased. |
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| 619. |
Increasing order of acidic character would beA. `CH_(3)COOH lt H_(2)SO_(4) lt H_(2)CO_(3)`B. `CH_(3)COOH lt H_(2)CO_(3) lt H_(2)SO_(4)`C. `H_(2)CO_(3) lt CH_(3)COOH lt H_(2)SO_(4)`D. `H_(2)SO_(4) lt H_(2)CO_(3) lt CH_(3)COOH` |
| Answer» Correct Answer - C | |
| 620. |
A solution of weak acids is diluted by adding an equal volume of water. Which of the following will not changeA. Strength of the acidB. The value of `[H_(3)O^(+)]`C. pH of the solutionD. The degree of dissociation of acid |
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Answer» Correct Answer - B The value of `H_(3)O^(+)` ions will not changed. `CH_(3)COOH + H_(2)O CH_(3)COO^(-) + H_(3)O^(-)`. |
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| 621. |
What is the pH for a neutral solutions at the normal temperature of the human bodyA. 7.2B. `14.0`C. 6.8D. `6.0` |
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Answer» Correct Answer - C Human body contain buffer solution. Its pH = 6.8. |
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| 622. |
What will be hydrogen ion concentration in moles `"litre"^(-1)` of a solution, whose pH is 4.58A. `2.63 xx 10^(-5)`B. `3.0 xx 10^(-5)`C. 4.68D. None of these |
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Answer» Correct Answer - A `[H^(+)] = "Antilog" (-4.58)` `[H^(+)] = 2. 63 xx 10^(-5)` moles/litre. |
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| 623. |
Calculate the `H^(+)` ion concentration in a 1.00(M) HCN litre solution `(K_(a) = 4 xx 10^(-10))`A. `4 xx 10^(-14)` mole/litreB. `2 xx 10^(-5)` mole/litreC. `2.5 xx 10^(-5)` mole/litreD. None of these |
| Answer» Correct Answer - B | |
| 624. |
Which of the following mixtures will act as buffer?A. `H_(2)CO_(3)+NaOH(1.5:1 "molar ratio")`B. `H_(2)CO_(3)+NaOH(1.5:2 "molar ratio")`C. `NH_(4)OH+HCl(5:4 "molar ratio")`D. `NH_(4)OH+HCl(4:5 "molar ratio")` |
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Answer» Correct Answer - a,b,c |
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| 625. |
Number of `H^(+)` ions present in 10 mL of solution of `pH=3` are:A. `10^(13)`B. `6.02xx10^(18)`C. `6.02xx10^(13)`D. `6.002xx10^(10)` |
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Answer» Correct Answer - B Meq.of `H^(+)= 10^(-3)xx10=10^(-2)` `:. "No. of" H^(+)= (10^(-2)xx6.02xx10^(23))/(1000)` `=6.02xx10^(18)` |
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| 626. |
Which of the following solutions can act as bufferA. 0.1 molar aq. NaClB. 0.1 molar aq. `CH_(3)COOH + 0.1` molar NaOHC. 0.1 molar aq. Ammonium acetateD. None of the above |
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Answer» Correct Answer - C `CH_(3)COONH_(4)` is a simple buffer and called salt of weak acid and weak base. |
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| 627. |
If the hydrogen ion concentration of a given solution is `5.5 xx 10^(-3) " mol litre"^(-1)`, the pH of the solution will beA. 2.26B. 3.4C. 3.75D. 2.76 |
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Answer» Correct Answer - A `[H^(+)] = 5.5 xx 10^(-3)` mole/litre `pH = -log [H^(+)] , pH = -log [ 5.5 xx 10^(-3)] , pH = 2.26`. |
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| 628. |
Which of the following solutions can act as buffer?A. `0.1` molar aq.`NaCl`B. `0.1` molar aq.`CH_(3)COOH + 0.1` molar `NaOH`C. `0.1` molar aq.Ammonium acetateD. `0.1` molar `H_(3)PO_(4)` |
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Answer» Correct Answer - C Salt of weak acid and weak base acts as buffer |
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| 629. |
Which of the following solution cannot act as buffer?A. `NaH_(2)PO_(4) + H_(3)PO_(4)`B. `CH_(3)COOH + CH_(3)COONa`C. `HCl + NH_(4)Cl`D. `H_(3)PO_(4) + Na_(2)HPO_(4)` |
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Answer» Correct Answer - C Mixture of `1`. W.A and its salt of SB & 2. W.B and it salt of strong acid acts as a buffer |
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| 630. |
Which of the following solutions cannot act as a bufferA. `NaH_(2)PO_(4)+H_(3)PO_(4)`B. `CH_(3)COOH + CH_(3)COONa`C. `HCl+NH_(4)Cl`D. `H_(3)PO_(4) + Na_(2)HPO_(4)` |
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Answer» Correct Answer - C A strong acid is not used to make a bufer. |
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| 631. |
The pH of a solution is increased from 3 to 6. Its `H^(+)` ion concentration will beA. Reduced to halfB. DoubledC. Reduced by 1000 timesD. increased by 1000 times |
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Answer» Correct Answer - C When pH = 3, then `[H^(+)] = 10^(-3) M` after that we increased pH from 3 to 6 then `[H^(+)] = 10^(-6)M` means reduced 1000 times. |
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| 632. |
If `H_(3)O^(+)` ion concentration of a solution is increased by 10 times , its pH willA. Increase by `1`B. Remains unchangedC. Decreases by `1`D. Increase by `10` |
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Answer» Correct Answer - C Let the initial concentration of `H^(o+) = 10^(-1)M` `pH = 1` The new concnetration of `H^(o+)`, is `10` times the initial concentration of `H^(o+) = 10 xx 10^(-1) M = 10^(@) = 1 pH_(2) =- log(1) = 0` The change in `pH = pH_(1) - pH_(2) = 1.0 = 0` Change in `pH` is one and it decreases from `1` to zero. |
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| 633. |
Which of the following curves corresponds to the titration of a weak base with a strong acid?A. B. C. D. |
| Answer» Correct Answer - C | |
| 634. |
The amount of sodium hydrogen carbonate, `NaHCO_(3)`, in an antacid tablet is to be determined by dissolving the tablet in water and titrating the resulting solution with hydrochloric acid. Which indicator is the most appropriate for this titration ? `{:("Acid", " "K_(a)),(H_(2)CO_(3),2.5xx10^(-4)),(HCO^(-3),2.5xx10^(-8)):}`A. methyl orange, `pK_(In)=3.7`B. bromothymol blue, `pK_(In)=7.0`C. phenolphthalein,`pK_(In)=9.3`D. alizarin yellow `pK_(In)=12.5` |
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Answer» Correct Answer - a |
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| 635. |
The pH of `0.001 M HCN` isA. `3`B. `11`C. Between `3` & `7`D. `7` |
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Answer» Correct Answer - C `[H^(+)] = 10^(-3)` `P^(H) = -log_(10)[H^(+)]` `HCN` is weak acid. So `P^(H)` value between `3` & `7` |
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| 636. |
What is the pH at the equivalenve point for the titration of `0.20 M` aniline `(K_(b) = 7.95 xx 10^(-10))` with `0.20 M HCL`A. `2.8`B. `11.1`C. `5.2`D. `9.42` |
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Answer» Correct Answer - A At the equivalence point in the titration of weak base and strong acid, it is the salt that is formed which undergoes hydrolysis `:. pH = 7 - (1)/(2) pK_(b) - 1/2 "log"C` |
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| 637. |
The pH value of `1.0 xx 10^(-8) M HCl` solution is less than 8 becauseA. HCl is completely ionised at this concentrationB. the ionization of water is negligibleC. The ionization of water cannot be assumed to be negligible in comparison with this low concentration of HClD. The pH cannot be calculated at such a low concentraion of HCl |
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Answer» Correct Answer - C As the solution is acidic, `pH lt 7`. This is because `[H^(+)]`. From `H_(2)O` cannot be neglected in comparison to `10^(-8)`. |
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| 638. |
To obtain a buffer wich should be suitable for maintaining a pH of about 4-5, we need to have in solution, a mixture ofA. A strong base + its salt with a weak acidB. A weak base + its salt with a strong acidC. A strong acid + its salt with a weak baseD. A weak acid + its salt with a strong base |
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Answer» Correct Answer - D A weak acid and its salt with a strong base maintain pH 4 - 5. |
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| 639. |
Which solution contains maximum number of `H^(+)` ionA. 0.1 M HClB. `0.1 M NH_(4)Cl`C. `0.1 M NaHCO_(3)`D. `0.1 M CH_(3)COOH` |
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Answer» Correct Answer - A The HCl is a strong acid and they lose easily `H^(+)` in solution. |
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| 640. |
Which of the following does not make any change in pH when added to 10 mL dilute HClA. 5 ml pure waterB. 20 ml pure waterC. 10 ml HClD. Same 20 ml dilute HCl |
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Answer» Correct Answer - D Buffer solution is formed. So the pH will not change. |
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| 641. |
The pH of 0.1 M acetic acid is 3, the dissociation constant of acid will beA. `1.0 xx 10^(-4)`B. `1.0 xx 10^(-5)`C. `1.0 xx 10^(-3)`D. `1.0 xx 10^(-8)` |
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Answer» Correct Answer - B `pH = 3, [H^(+)] = 10^(-3)M` `because [H^(+)] = sqrt(K xx c)` `[10^(-3)]^(2) = K xx c, ([10^(-6)])/(0.1) = K = 10^(-5)`. |
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| 642. |
Total number of moles for the reaction `2Hl rarr H_(2) + I_(2)`. If `alpha` is degree of dissociation isA. 2B. `2 - alpha`C. 1D. `1 - alpha` |
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Answer» Correct Answer - A `{:("Initial",2HI,hArr,H_(2)+,I_(2)),(,,,0,0):}` At equilibrium `2 - alpha (alpha)/(2) (alpha)/(2)` Total mole `= 2 - alpha + (alpha)/(2) + (alpha)/(2) = 2`. |
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| 643. |
Assertion : pH of hydrochloric acid solution is less than that of acetic acid solution of the same concentration . Reason : In equimolar solutions, the number of titrable protons present in hydrochloric acid is less than that present in acetic acid.A. If both assertion and reason are true and reason is the correct explanation of the assertionB. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false. |
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Answer» Correct Answer - C HCl is a strong electrolyte since it will produce more `H^(+)`, comparison than that of `CH_(3)COOH`. Hence assertion is true but reason false. |
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| 644. |
For a weak acid HA with dissociation constant `10^(-9)`, pOH of its 0.1 M solutions isA. 9B. 3C. 11D. 10 |
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Answer» Correct Answer - D Dissociation constant of `HA = 10^(-9)` `HA hArr H^(+) + A^(-)` `[H^(+)] = sqrt((K_(a))/(C)) = sqrt((10^(-9))/(0.1)) , [H^(+)] = 10^(-4)` `:. pH = 4` `because pH + pOH = 14` `pOH = 14- pH = 14-4 , pOH = 10` . |
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| 645. |
Assertion : NaCl is precipitated when HCl gas is passed in a saturated solution of NaCl. Reason : HCl is strong acid.A. If both assertion and reason are true and reason is the correct explanation of the assertionB. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false. |
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Answer» Correct Answer - B Pure NaCl is precipitated when HCl gas is passed in a saturated solution of NaCl because the value of `[Na^(+)]` and `[Cl^(-)]` becomes greater than `K_(sp)` of NaCl. |
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| 646. |
Let the solubilities of `AgCl` in pure water be `0.01 M CaCl_(2)`, `0.01 M NaCl` and `0.05 M AgNO_(3)` be `s_(1)`, `s_(2)`, `s_(3)` and `s_(4)` respectively. What is the correct order of these quantities? Neglect any complexation.A. `s_(1) gt s_(2) gt s_(3) gt s_(4)`B. `s_(1) gt s_(2) = s_(3) gt s_(4)`C. `s_(1) gt s_(3) gt s_(2) gt s_(4)`D. `s_(4) gt s_(2) gt s_(3) gt s_(1)` |
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Answer» Correct Answer - c |
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| 647. |
What is the correct relationship between the pHs of isomolar solutions of sodium oxide `(pH_(1))`, sodium sulphide `(pH_(2))`, sodium selenide `(pH_(3))` and sodium telluride `(pH_(4))`A. `pH_(1) gt pH_(2) = pH_(3) gt pH_(4)`B. `pH_(1) lt pH_(2) lt pH_(3) lt pH_(4)`C. `pH_(1) lt pH_(2) lt pH_(3) = pH_(4)`D. `pH_(1) gt pH_(2) gt pH_(3) gt pH_(4)` |
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Answer» Correct Answer - D Order of acidic strength is `H_(2)Te gt H_(2)Se gt H_(2)S gt H_(2)O` `Na_(2)O` is a salt of `NaOH + H_(2)O` and `H_(2)S` is least acidic among given acids hence pH in this case will be `"max"^(m)`. |
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| 648. |
`H_(3)BO_(3)` is.A. monobasic and weak Lewis acidB. monobasic and weak Bronsted acidC. monobasic and strong Lewis acidD. tribasic and weak Bronsted acid |
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Answer» Correct Answer - A `H_(3)BO_(3)+OH^(-)rarrB(OH)_(4)^(-)+H^(+)` Accepts lone pair of electron thus Lewis acid but weak. |
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| 649. |
The specific gravity of `H_(2)SO_(4)` is `1.8 g// c c` and this solution is found to contain `98% H_(2)SO_(4)` by weight. `10c c` of this solution is mixed with `350 c c` of pure water. `25 mL `of this dil. `H_(2)SO_(4)` solution neutralises `500 mL` of `NaOH` solution. Then the `P^(H)` of `NaOH` solution isA. `12.398`B. `1.602`C. `12.699`D. `12.301` |
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Answer» Correct Answer - C Normaility of `H_(2)SO_(4) = ("Sp.gr" xx % xx 10)/("Ew.wt")` `= (1.8 xx 98 xx 10)/(49) = 36` on dilution with water, reultant Normality of `H_(2)SO_(4) = (10 xx 36)/(360) = 1N` `H_(2)SO_(4) + NaOH rarr Na_(2)SO_(4) + 2H_(2)O` for neutralisation no. of mili equivalents of acid and base must be equal then `N _(H_(2)SO_(4))V_(N_(2)SO_(4)) = N_(NaOH) V_(NaOH)` `1 xx 25 = N_(NaOH) xx 500` `N_(NaOH) = 1/20 = 0.5 = 5 xx 10^(-2)` `P^(OH) = 2 - "log" 5 = 2 - 0.6990` `p^(H)` of `NaOH = 12.6990` |
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| 650. |
`pH` calculation upon dilute of strong acid solution is generally done by equating `n_(H)` in original solution & diluted solution.However . If strong acid solution is very dilute then `H^(+)` from water are also to be considered take `log3.7=0.568` and answer the following questions. A `1` litre solution of `pH=6` (solution of a strong acid) is added to the `7//3` litres of water.What is the `pH` of resulting solution? Neglect the common ion effect on `H_(2)O`.A. `6.4`B. `6.52`C. `6.365`D. `6.432` |
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Answer» `pH=6` `[H^(+)]=10^(-6)` `N_(1)V_(1)=N_(1)V_(2)` rArr `10^(-6)xx1=N_(2)[1+(7)/(3)]" "rArr" "10^(-6)=N_(2)xx(10)/(3)` `N_(2)=(3)/(10)xx10^(-6)" "rArr " "N_(2)=3xx10^(-7)` `[H^(+)]lt10^(-6)` so `[H^(+)]` of water is also added as common ion effect on `H_(2)O` is neglected so `[H^(+)]=3xx10^(-7)+10^(-7)=4xx10^(-7)M` `rArr pH=7-log4=7-0.60=6.4` |
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