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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.
| 701. |
Equal volumes of following solutions are mixed. In which case the pH of resulting solution will be average value of pH of two solutions?A. `pH=2(HCl)and pH=12(NaOH)`B. `pH=2(HCl)and pH=4(HCl)`C. `pH=2(HCN)and pH=12(NaOH)(K_(a)of HCN=10^(-10))`D. `pH=5(CH_(3)COOH)and pH=9(NH_(3))(aq)(K_(a)of HCN=10^(-10))` |
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Answer» Correct Answer - A::D |
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| 702. |
Degree of dissociation `(alpha)` `alpha` are the number of moles which are dissociating from 1 mole of given reactants and gas density measurements can be used to determine the degree of dissociatin. Let us take a general case where one molecule of a substance A splits up into n molecules of A(g) on heating i.e., `A_(n)(g)hArrnA(g)` `t=0a` `t =t_(eq)a-x nx " "alpha=x/aimpliesx=a alpha` `a-a alpha n a alpha` Total number of Moles `=a-a alpha +n a alpha``=[1+(n-1)alpha]a` Observed molecular weight of molar mass of the mixture `M_("mixture")=(M_(A_(n)))/([1+(n-1)alpha]),M_(A_(n))=` Molar mass of `A_(n)` If the t otal mass of the mixture in question (1) is 300 gm, then moles of C(g) present areA. `1/4`B. `4/3`C. `3/4`D. `1/2` |
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Answer» Correct Answer - C |
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| 703. |
Name a species which can act both as conjugate acid and conjugate base. |
| Answer» Correct Answer - `H_(2)O` | |
| 704. |
Degree of dissociation `(alpha)` `alpha` are the number of moles which are dissociating from 1 mole of given reactants and gas density measurements can be used to determine the degree of dissociatin. Let us take a general case where one molecule of a substance A splits up into n molecules of A(g) on heating i.e., `A_(n)(g)hArrnA(g)` `t=0a` `t =t_(eq)a-x nx " "alpha=x/aimpliesx=a alpha` `a-a alpha n a alpha` Total number of Moles `=a-a alpha +n a alpha``=[1+(n-1)alpha]a` Observed molecular weight of molar mass of the mixture `M_("mixture")=(M_(A_(n)))/([1+(n-1)alpha]),M_(A_(n))=` Molar mass of `A_(n)` A sample of mixture A(g), B(g)and C(g) under equlibrium has a mean molecular weight (observed) of 80. The equlibrium is `A(g)hArrB(g)+C(g)` (Mol wt =100)`" "` (Mol. wt=60)`" "` (Mol. wt=40)Calculate the Degree of dissociation for given reaction.A. `0.25`B. `0.5`C. `0.75`D. `0.8` |
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Answer» Correct Answer - A |
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| 705. |
Which of the following solutions will have pH close to 1.0 ?A. 100 ml of `(M//10)` HCI + 100 ml of `(M//10)` NaOHB. 55 ml of `(M//10)` HCI+ 45 ml of `(M//10)` NaOhC. 10ml of `(M//10)` HCI + 90 ml of `(M//10)` NaOND. 75 ml of `(M//10)` HCI + 25 ml of `(M//5)` NaOH |
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Answer» Correct Answer - D ` "In " (d) M_(H^(+)) =(M_(1)V_(1)- M_(2)V_(2))/(V_(1)+V_(2)) ` `=(0.1 xx 75 -0.2xx 25)/(100) = (7.5 -5.0)/(100)` `M_(H^(+)) =(2.5)/(100) =2.5 xx 10^(-2) M = 25 xx 10^(-3) M` `pH =- log [H^(+)] =- log (25xx10^(-3))` `=- 1.398 +3 = 1.602` In this case pH is close to 1. |
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| 706. |
We use a number of acids and bases in our life. Some of these are used as medicine or in household whereas some others are used in the industry. Besides, many fruits, vegetables and other eatables also contain acids, useful for the living system. Now answer the following questions: (i) Name three acids used in the eatables. Give one use of them. (ii) Name three acids used in the industry. Give one important use of each of them. (iii) Name five fruits or vegetables and the acid present in each of them. (iv) Name three bases used in the industry or in household or as medicine and give one important use of each of them. |
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Answer» (i) Benzoic acid to preserve food, Carbonic acid to make aerated drinks and ethanoic acid used in vinegar. (ii) HCl to clean metals before electroplating or household cleaning, nitric acid in the production of fertilizers and explosives and sulphuric acid as a battery acid and in making fertilizers. (iii) Apples-malic acid, Orange-citric acid, Bananas-malic acid and citric acid, grapesmalic acid and tartaric acid(3:2), Potatoes-malic acid and citric acid, Tomatoes-citric acid and malic acid. (iv) Ammonia in the manufacture of fertilizers, NaOH in the manufacture of soaps, detergents and cleaners and Magnesium hydroxide as antacid to neutralize acidity in stomach. |
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| 707. |
Which of the following solution will have pH close to 7?A. `100mlof M/10HCl+100ml of M/10NaOH`B. 1M solution of `CH_(3)COONH_(4)(K_(a)=K_(b))`C. `1Mml of M/10H_(2)SO_(4)+100mlof M/10NaOH`D. `100ml of M/10HCl+100mlof M/10Ca(OH)_(2)` |
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Answer» Correct Answer - A::B |
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| 708. |
What is the difference in `pH` for `1//3` and `2//3` stages of neutralization of `0.1 M CH_(3)COOH` with `0.1 M NaOH`?A. `2log""(1)/(4)`B. `2 log 3`C. `0.9542`D. `0.3010` |
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Answer» Correct Answer - B::C |
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| 709. |
Calculate the pH of each of the following solution (i) 100 ml of 0.1 M `CH_(3)COOH` mixed with 100 ml of 0.1 M NaOH. (ii) 100 ml of 0.1 M `CH_(3)COOH` mixed with 50 ml of 0.1 m NaOH (iii) `50 ml of 0.1 M CH_(3)COOH` mixed with 100 ml of 0.1 M NaOH. `K_(a)(CH_(3)COOH)=1.8xx10^(-5)` |
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Answer» (a) `underset("1 mole")(CH_(3)COOH)+underset("1 mole")(NaOH)toCH_(3)underset("1 mole")(COONa)+underset("1 mole")(H_(2)O)` m. moles of `Ch_(3)COOH=0.1xx100=10` m moles of `NaOH=100xx0.1=10` Therefore, m. moles of `CH_(3)COONa` is 10 Now in the solution only salt `CH_(3)COONa` is present, therefore, pH of solution can be calculated by the hydrolysis of `CH_(3)COONa` `CH_(3)COONa` is a salt of weak acid with strong base `pH=7+1/2pK_(a)+1/2logC` `=7+1/2xx4.75+1/2log((10)/(200))" "(pK_(a)=-log1.8xx10^(-5))` `=8.72" "pK_(a)=4.75` (b) `CH_(3)COOH+NaOHtoCH_(3)COONa+H_(2)O` `{:("Initial m moles",10,5,0,0),("m moles after reaction",10-5,0,5,5):}` Since in the solution, mixture of `CH_(3)COOH and CH_(3)COONa` is present therefore, solution is an acidic buffer and pH can be calculated as `pH=pK_(a)+log""(["salt"])/(["acid"])pH=4.75+log""(((5)/(200))/((5)/(200)))pH=4.75` (c ) `CH_(3)COOH+NaOHto CH_(3)COONa+H_(2)O` `{:("Initial m moles",5,10,0,0),("m moles after reaction",0,(10-5),5,5):}` Now in the solution strong base NaOH and salt of weak aid with strong base is present We can assume that `OH^(-)` ions concentration from the hydrolysis of salt is negligible because hydrlysis of `CH_(3)COONa` is suppressed by the presence of strong base NaOH. `underset(("from salt"))(CH_(3)COO^(-))+H_(2)OhArrCH_(3)COOH+OH^(-)` `therefore[OH^(-)]=([5])/([150])=((1)/(30))` `pOH=log[OH^(-)]=-log((1)/(30))=log30=1.4771` `pH+pOH=14` `pH=14-1.4771=12.52` |
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| 710. |
If the solution `0.5 M NH_(3)` and stability constant for `A^(+)(NH_(3))_(2)` is `K_(stb)=([Ag(NH_(3))_(2)]^(+))/([Ag^(+)(aq)][NH_(3)]^(2))=6.4xx10^(7),` then find the solubility of AgCl in the above solution `K_(sp)of AgCl=2xx10^(-10)` |
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Answer» `K_(eq)=K_(sp)xxK_(stb)` `implies2xx10^(-10)xx6.4xx10^(7)=([Ag(NH_(3))_(2)][Cl^(-)])/([NH_(3)]^(2))=(SxxS)/((0.5-2S)^(2))` `implies(S)/((0.5-2S))=sqrt(2xx64xx10^(-4))` `impliesS=0.046` |
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| 711. |
STATEMENT-1 : When a small amount of strong acid is added to a buffer solution, its pH value does not change signific antly STATEMENT-2 : Buffer action of the buffer sloution resist the changee in pH when small amount of acid is added to itA. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-4B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-4C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
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Answer» Correct Answer - A |
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| 712. |
Calculate [OH–] if pOH = 8.3. |
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Answer» pOH = -log10[OH–]; 8.3 = -log10[OH–] Taking antilog on both the sides [OH–] = antilog (-8.3) = [OH–] = antilog (-9 - 8.3 + 9) = antilog [(0.7) × 10-9] = 5.012 × 10-9 mol/dm3 |
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| 713. |
What is the effect of increase of temperature on the pH of a Buffer solution? |
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Answer» pH of a buffer changes with temperature because Kw changes, pH decreases with increase of temperature. |
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| 714. |
Calculate [H+] if pOH = 9.23 |
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Answer» pH + pOH = 14 pH = 14 – pOH = 14 – 9.23 = 4.77 pH = -log10[H+] = -4.77 = log[H+] Taking antilog on both the sides [H+] = antilog (-4.77) = antilog [+5 -4.77 – 5] = antilog [0.23] × 10-5 = 1.698 × 10-5] mol/dm3] |
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| 715. |
The pOH of a solution is 5.725. Calculate the [H+]. |
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Answer» pH + pOH = 14 pH = 14 – pOH = 14 – 5.725 = 8.275 pH = -log10[H+] ⇒ 8.275 = log10[H+] ⇒ [H+] = antilog (-8.275) = antilog(9 – 8.275 – 9) [H+] = antilog (0.275 – 9) = 5.309 × 10-9 mol dm-3. |
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| 716. |
Explain Mechanism or working of acid buffer. |
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Answer» Consider acidic buffer mixture, CH3COOH ⇌ CH3COO– + H+; CH3COONa → CH3COO– + Na+ Here, acetic acid is a weak electrolyte, in it’s solution there exists equilibrium between it’s ions and molecules. Where as sodium acetate completely dissociates into it’s ions. Therefore, the buffer mixture contains large number of CH3COO– ions followed by Na+ , H+ , CH3COOH. Case i) When an acid added to this solution: H+ ion of the acid combines with CH3COO– ion in buffer solution and makes equilibrium with acetic acid. CH3COO– + H+ ⇌ CH3COOH As a result pH remains constant. Case ii) When a base is added to this solution: OH– ion of the base combines with H+ ion in buffer solution forms water molecule. H+ + OH– → H2O As a result pH remains constant. |
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| 717. |
STATEMENT-1: pH of water decreases with increase in temperature. STATEMENT-2 : `K_(w)` of water decreases with increase in temperature.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-5B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-5C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
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Answer» Correct Answer - C |
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| 718. |
Give examples of Acidic and Basic Buffers? |
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Answer» Acidic Buffer – CH3COOH + CH3COONa Basic Buffer - NH4OH + NH4Cl |
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| 719. |
The units of solubility product of silver chromate `(AgCrO_(4))` will beA. `mol^(2)L^(-2)`B. `mol^(3)L^(-3)`C. `mol L^(-1)`D. `mol^(-1)L` |
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Answer» Correct Answer - B The solubility product is an equilibrium constant. `Ag_(2)CrO_(4)(s)hArr2Ag^(+)(aq.)+CrO_(4)^(2-)(aq.)` `K_(sp) = C_(Ag^(+))^(2)C_(CrO_(4)^(2-))` `= (mol L^(-1))^(2)(mol L^(-1))= mol^(3)L^(-3)` |
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| 720. |
Calculate the [OH] of a solution whose pH is 9.62. |
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Answer» pH + pOH = 14 pOH = 14 – pH = 14 – 9.62 = 4.38 pOH = 4.38 = -log10[OH–]; 4.38 = -log10[OH-] log10[OH–] = -4.38; [OH-] = antilog (-4.38) = antilog (5 – 4.38 – 5) = antilog (0.62 – 5) = antilog (0.62) × 10-5 = 4.169 × 10-5 mol dm-3 |
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| 721. |
The solubility proudct of AgCl at `25^(@)C` is `1xx10^(-10)` A solution of `Ag^(+)` at a concentration `4xx10^(-3)` M just fails to yield a prenciitate of AgCl with concentration of `1xx10^(-3) M Cl^(-)` when the concentration of `NH_(3)` in the solution is `2xx10^(-2)M.` Calculate the equlibrium constant for `[Ag(NH_(3))_(2))hArrAg^(+)+2NH_(3)` |
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Answer» `AgCl(s)hArrAg^(+)(aq)+Cl^(-)(aq)` `K_(sp)=[Ag^(+)][Cl^(-)]` `10^(-10)=[Ag^(+)][10^(-3)]` `[Ag^(+)]=10^(-7)M` Since the initial concentration of `Ag^(+)is 4xx10^(-3)M,` therefore, ionic product of AgCl will be `[Ag^(+)][Cl^(-)]=4xx10^(-3)xx10^(-3)=4xx10^(-6)` which is greater than `K_(sp)` Thus precipitation will take place but in the question at the concentration of `Ag^(+)` ion `4xx10^(-3)` M there is no precipitation, it means the concentration of `Ag^(+)` will be equal to `10^(-7)` M or less than this. But the precipitation just fails, therefore, we assume that the equlibrium concentration of `Ag^(+)is 10^(-7)` M. `Ag^(+)+2NH_(3)hArr[Ag(NH_(3))_(2)]^(+)` `{:("Initial conc.",4xx10^(-3)M,CM,0),("At equlibrium",(4xx10^(-3)-x),(c-2x)M,xM):}` Since at equlibrium `[Ag^(+)]=10^(-7)M` and conc. of `NH_(3)=2xx10^(-2)M` `4xx10^(-3)-x~~10^(-7)" "(C-2x)=2xx10^(-2)` `thereforex~~4xx10^(-3)` `K_(f)([Ag(NH_(3))_(2)]^(+))/([Ag^(+)][NH_(3)]^(2))=(4xx10^(-3))/((10^(-7))(2xx10^(-2)))=10^(5)` `therefore` Therefore equlibrium constant for the reaction `[Ag(NH_(3))_(2)]^(+)hArrAg^(+)+2NH_(3)K_(f)` `K_(f)=(1)/(K_(f))=(1)/(10^(8))=10^(-8)` |
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| 722. |
Give the Henderson's - Hasselbalch equation for an acidic buffer solution. |
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Answer» pH = pKa + log \(\frac{[Salt]}{[Acid]}\) |
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| 723. |
Statement-1. pH of water increases with increase in temperature. Statement-2. `K_(w)` of water increases with increase in temperature.A. Statement-1 is True, Statement-2 is True , Statement-2 is the correct explanation of Statement - 1 .B. Statement-1 is True, Statement - 2 is True , Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
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Answer» Correct Answer - D Correct statement-1. pH of water decreases with increase in temperature. |
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| 724. |
Give the Henderson's-Hasselbalch equation for a basic buffer solution. |
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Answer» pOH = pKb + log \(\frac{[Salt]}{[Acid]}\) |
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| 725. |
Which of the following is the correct representation of the solubility product expression for mercurous iodide `(HgI_(2))` ?A. `[Hg_(2)^(2+)][I^(-)]^(2)`B. `[Hg^(+)]^(2)[I^(-)]^(2)`C. `[Hg^(+)][I^(-)]`D. `[Hg^(2+)][I^(-)]^(2)` |
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Answer» Correct Answer - A The solubility equilibrium can be represented as `Hg_(2)I_(2)(s)hArrHg_(2)^(2+)(aq.)+2I^(-)(aq.)` The solubility product of a compound is the product of the molar concentrations of the constituent ions, each raised to the power of its stoichiometric coefficient in the equilibrium equation: `K_(sp)=[Hg_(2)^(2+)][I^(-)]^(2)` |
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| 726. |
Statement -1 `HCO_(3)^(-)` ion can act as a strong base. Statement -2 `CO_(3)^(2-)` ion can act as a weak base.A. Statements -1 is true , statement -2 is also true, statement -2 is the correct explanation of statement -1B. Statement -1 is true , statement-2 is also true, statement-2 is not the correct explanation of statement-1C. Statement -1 is true, statement -2 is false.D. Statement -1 is false, satement-2 is true. |
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Answer» Correct Answer - B Correct explanation . `HCO_(3)^(-)` ion can act as a strong base because the conjugate acid `H_(2)CO_(3)` is a weak acid `underset(("Strong"))(HCO_(3)^(-)) +H^(+) hArr underset(("Weak"))(H_(2)CO_(3))` |
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| 727. |
3.2 moles of HI (g) were heated in a sealed bulb at `444^(@)C` till the equlibrium was reached its degree of dissociation was found to be `20%` Calculate the number of moles of hydrogen iodide, hydrogen and iodine present at eth equlibrium point and determine the value of equlibrium constnat for the reaction `2Hl(g)hArrH_(2)(g)+I_(2)(g).` Considering the volume of the container 1 L. |
| Answer» `K_(c)=0.0156,[Hl]=2.56M,[H_(2)]=[l_(2)]=0.32M` | |
| 728. |
Define acidolysis. |
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Answer» Acidolysis is the process where acidity of blood decreases from 7.3. |
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| 729. |
Name a basic buffer having pH around 10. |
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Answer» Basic buffer Na2B4O7 + Na OH Borax sodium hydroxide. |
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| 730. |
Statement -1 Soda water becomes flat if kept open. Statement -2 the amount of dissolved carbon dioxide decrease with increase in pressure.A. Statements -1 is true , statement -2 is also true, statement -2 is the correct explanation of statement -3B. Statement -1 is true , statement-2 is also true, statement-2 is not the correct explanation of statement-3C. Statement -1 is true, statement -2 is false.D. Statement -1 is false, satement-2 is true. |
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Answer» Correct Answer - A Statement-2 is the correct explanation for statement-1. |
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| 731. |
Statement-1. `HCO_(3)^(-)` ion act as a strong acid as well as strong base. Statement-2. `CO_(2)^(2-)` ions act only as weak base.A. Statement-1 is True, Statement-2 is True , Statement-2 is the correct explanation of Statement - 1 .B. Statement-1 is True, Statement - 2 is True , Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
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Answer» Correct Answer - B Correct explanation. `CO_(3)^(2-)` ions act as weak base whereas `H_(2)CO_(3)` is a weak acid. |
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| 732. |
Define Alkolysis. |
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Answer» Alkolysis is the process where acidity of blood increases from 8. |
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| 733. |
At 500 K, equlibrium constant, `K_(c)` for the following reaction is 5. `1//2 H_(2)(g)+ 1//2(g)hArr HI (g)` What would be the equilibrium constant `K_(c)` for the reaction `2hi(g)hArrH_(2)(g)+l_(2)(g)`A. 0.04B. 0.4C. 25D. 2.5 |
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Answer» Correct Answer - A For the reaction, `1//2H_(2)(g)+1//2 I_(2)(g)hArrHI(g),K_(c)=5` For the reaction, `HI(g)hArr1//2H_(2)(g)+1//2l_(2)(g),K_(c)=(1)/(5)` For the reaction `2HI(g)hArrH_(2)(g)+l_(2)(g),K_(c)=((1)/(5))^(1//2)` |
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| 734. |
The pH of blood circulating in a human body is maintained around. `7.4` by the action of the buffer systemA. `CH_(3)COOH//CH_(3)COONa`B. `NH_(4)CI//NH_(3)`C. `H_(2)PO_(4)^(-)//HPO_(4)^(2-)`D. `CO_(2)//HCO_(3)^(-)` |
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Answer» Correct Answer - D Blood, as a buffer solution, contains `H_(2)CO_(3)` as well as other conjugate acid-base pairs. Carbon dioxide might not appear to be an acid, since it has no proton of it froms carbonic acid `(H_(2)CO_(3))`: `CO_(2)(aq.)+H_(2)hArrH_(2)CO_(3)(aq.)` The conjugate base of carbonic acid is the bicarbonate ion, `HCO_(3)^(-)`: `H_(2)CO_(3)(aq.)hArrHCO_(3)^(-)(aq.)+H^(+)(aq.)` The overall reaction relating carbon dioxide and bicarbonate ion is the sum of these two reactions: `CO_(2)(aq.)+H_(2)OhArrHCO_(3)^(-)(aq.)+H^(+)(aq.)` The `pK_(a)` of the equilibrium for this buffer system is `6.37`. A second class of physiological buffer is based on the dihydrogen phosphate anion `(H_(2)PO_(4)^(-))` which is a weak acid and hydrogen phosphate `(HPO_(4)^(2-))`, its conjugate base: `H_(2)PO_(4)^(-)(aq.)hArrHPO_(4)^(2-)(aq.)+H^(+)(aq.)` Again the `pK_(a)(7.21)` is close to the physiological pH. |
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| 735. |
What do you mean by buffer solution? |
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Answer» The solution which resist the pH or pOH when an acid or base is added to it. Buffer solution is a mixture of weak electrolyte and its salt. |
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| 736. |
Statement -1 Reaction between iron and steam is irreversible if carried in an open container. Statement -2 An irreversible reaction cannot proceed in the reverse direction.A. Statements -1 is true , statement -2 is also true, statement -2 is the correct explanation of statement -4B. Statement -1 is true , statement-2 is also true, statement-2 is not the correct explanation of statement-4C. Statement -1 is true, statement -2 is false.D. Statement -1 is false, satement-2 is true. |
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Answer» Correct Answer - B Correct explanation. The gaseous species , both steam and hydrogen escape out of the container. |
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| 737. |
Define buffer action. |
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Answer» The ability or capacity of a buffer solution to resist the pH by addition of acid or base. |
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| 738. |
Define Reversible reaction with example. |
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Answer» A Chemical reaction in which an infinitesimally small change in any one of the variable (Pressure, Temperature, Concentration of reactant or product) of the reaction is changed the direction (forward or backward) of the reaction changes. It is represented by half arrow mark. An Example is H2(g) +I2(g) ⇌ 2HI(g) |
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| 739. |
Define Irreversible reaction with example. |
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Answer» A Chemical reaction in which product can not able to give back the reactant are called Irreversible reaction. Example:- C(s) + O2(g) → CO2(g) |
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| 740. |
Mention the types of buffer solution. |
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Answer» 1. acidic buffer 2. basic buffer 3. neutral buffer. |
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| 741. |
500 mL of 0.2 M aqueous solution of acetic acid is mixed with 500 mL of 0.2 M HCl at `25^(@)C`. (i) Calculate the degree of dissociation of acetic acid in the resulting solution and pH of the solution. (ii) If 6 g of NaOH is added to the above solution, determine the final pH [Assume there is no change in volume on mixing : `K_(a) ` of acetic acid is `1.75xx10^(-5) " mol " L^(-1)`] |
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Answer» (i) Millimoles of `CH_(3)CO OH = 500 xx 0.2 = 100` Millimoles of HCl `= 500xx 0.2=100` Final volume after mixing `= 500 + 500 = 1000 ` mL `:. [ CH_(3)CO OH]=(100)/(1000) = 0.1 M, " " [HCl ] = (100)/(1000) = 0.1 M` `{:(,CH_(3)CO OH ,hArr,CH_(3)CO O^(-),+,H^(+),,),("Before dissociation",0.1 M ,," "0,,0.1 M,,),(,,,,,("from HCl"),,),("After dissociation",(0.1-x),," "x,,(0.1+x),,):}` `K_(a) = (x(0.1+))/((0.1+x))` As in the presence of HCl, dissociation of `CH_(3)CO OH` will be very small (due to common ion effect ), x is very very small. Hence, `K_(a) = (x(0.1))/(0.1) = x = 1.75xx10^(-5) ` mol `L^(-1) ` (Given) `:.` Degree of dissociation `= (x)/(0.1) = (1.75xx10^(-5))/(0.1) = 1.75xx10^(-4)= 0.00175 %` Further, `[H^(+)]= 0.1 + x ~~ 0.1 :. pH = - log 0.1 = 1 ` (ii) 6g of NaOH = `(6)/(40) ` mole = 0.15 mole Hence, now the equilibrium will be `{:(,CH_(3)CO OH ,+,HCl,+,NaOH,hArr,CH_(3)CO ONa ,+,NaCl,+,H_(2)O),("Initial",0.1,,0.1,,0.15,,0,,0,,0),("At eqm.",0.05,,0,,0,,0.05,,0,,0):}` Thus, the solution will now be 0.05 M in `CH_(3)CO OH` and 0.05 M in `CH_(3)CO ONa`, i.e., it is acidic buffer. `pH = - log K_(a) + log. (["Salt"])/(["Acid"]) = - log (1.75xx10^(-5)) + log (0.05)/(0.05) = 4. 757` |
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| 742. |
A solution give the following colors with different indicators. Methyl orange – yellow, methyl red – yellow, and bromothymol blue Orange . what is the pH of the solution? |
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Answer» (i) The colors in methyl orange indicates that pH > 4.5 (ii) Colors in methyl red indicates that pH > 6.0 and (iii) colors in bromothymol blue indicates that pH < 6.3. Therefore, the pH of the solution is between 6.0 to 6.3. |
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| 743. |
Mention the uses of buffer solution. |
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Answer» It is used in chemical industries like paper, sugar, pharmaceuticals. It is used in all biological fluids to maintain constant pH by buffer action. |
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| 744. |
Calculate the pH of an aqueous solution of 1.0 M ammonium formate assuming complete dissociation (`pK_(a)` of formic acid = 3.8 and `pK_(b)` of ammonia = 4.8 ). |
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Answer» Correct Answer - `6.5` Similar to Hint 6 (`HCO ONH_(4)` is a salt of weak acid, weak base). |
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| 745. |
0.03 mole of `Ca^(2+)` ions is added to a litre of 0.01 M `SO_(4)^(2-)` solution. Will it cause precipitation of `CaSO_(4)` ? `K_(sp) "for" CaSO_(4)=2.4xx10^(-5)`. |
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Answer» Correct Answer - Precipitation of `CaSO_(4)` will occur In the final solution, `[Ca^(2+)]=0.03 M and [SO_(4)^(2-)]=0.01 M`. Hence, ionic product of `CaSO_(4) = (0.03) (0.01)=3xx10^(-4)` which is greater than `K_(sp)`. Hence, precipitation will occur. |
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| 746. |
Define pH. Give its mathematical expression. |
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Answer» pH is defined as negative logarithm of [H2O+]concentration. |
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| 747. |
What is meant by ionic product of water (Ka)? |
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Answer» It is product of concentration of [H3O+] and [OH-] at a specific temperature. Kw = [H3O+][OH-] = 1.0 × 10-14 at 298K |
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| 748. |
What is the range of a pH indicator in terms of its dissociation constant `(K_(ln))`? |
| Answer» Correct Answer - `pH=pK _(ln) pm 1`. | |
| 749. |
Universal indicator shows green colour when pH of the solution is nearly....... . |
| Answer» Correct Answer - 8 | |
| 750. |
Why the Equilibrium constant is taken as dimensionless? |
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Answer» Activities are used in place of molar Conc. These represent the concentration or pressure with respect to standard state. |
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