DPP-47 to 48 With Answer Physical chemistry

DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 47 to 48 Class : XIII Course : DPP No.1 Total Marks : 45 Max. Time : 45 min. Single choice Objective ('–1' negative marking) Q.1 to Q.3 (3 marks 3 min.) [9, 9] Multiple choice objective ('–1' negative marking) Q.4 (4 marks 4 min.) [4, 4] Subjective Questions ('–1' negative marking) Q.5 to Q.12 (4 marks 4 min.) [32, 32] 1. Calculate the change in pH when a 0.1 M solution of CH3COOH in water at 25ºC is diluted to a final concentration of 0.01 M. [Ka = 1.85 × 10–5] tc 25ºCrki ij CH3COOHds 0.1Mfoy;u dks ty }kjk 0.01MlkUnzrk rd ruq fd;k tkrk gSA rks pHifjorZu dh x.kuk dhft,[K = 1.85 × 10–5] (1*) +0.5 (2) +0.4 (3) +0.7 (4) +0.6 Sol. pH = 1 {pK 2 a – log C} so (pH) = 1 [log C – log C ] = 1 {–1 + 2} = 0.5 [pH will increase]. 2 i f 2 2. At 25ºC , the dissociation constants of CH3COOH and NH4OH in an aqueous solution are almost the same. The pH of a solution of 0.01 N CH3COOH is 4 at 25ºC. The pH of 0.01 N NH4OH solution at the same temperature would be : 25ºCij,d tyh; foy;u esaCH3COOHrFkkNH4OHds fo;kstufLFkjkad dkeku yxHkxleku gSSA25ºCij0.01N CH3COOH ds foy;u dh pH 4gSA leku rkieku ij0.01N NH4OH foy;u dhpHD;k gksxh \ (1) 4 (2) 3 (3*) 10 (4) 11 3. The acid dissociation constants of H S and HS– are 10–7 and 10–13 respectively. The pH of 0.1M aqueous solution of H2S will be : H SrFkkHS– vEyds fo;kstufLFkjkaddkeku Øe'k%10–7 rFkk10–13 gSAH S ds0.1Mtyh;fo;yudhpHD;kgksxh\ 2 2 (1) 2 (2) 2 (3*) 4 (4) 5 4. Which of the following increase with dilution at a given temperature? (1*) pH of 103M acetic acid solution (2) pH of 103 M aniline solution (3*) degree of dissociation of 103M acetic acid (4*) degree of dissociation of 10–3M aniline fn;sx,rkiekuijfuEuesalsfdlesaruqrkdslkFko`f)gksrhgSA (1*) 103M ,lhfVd vEy ds foy;u dh pH (2) 103 M ,uhyhu ds foy;u dh pH (3*)103M,lhfVdvEy dsfo;kstu dhek=kk (4*)10–3M,uhyhudsfo;kstudhek=kk Sol. Degree of dissocation of WA & WB will increase. [H+] in WA and [OH–] in WB will decrease so pH of WA and pOH of WB will increase. 5. How much water must be added to 300 mL of a 0.2 M solution of CH3COOH for the degree of dissociation of the acid to double ? (Assume Ka of acetic acid is of order of 10 M) –5 CH3COOHds0.2Mfoy;uds 300mlesa fdruk ty feyk;k tk,]fd vEyds fo;kstudh ek=kknks xquhgks tk, ¼;g ekudj dh ,lhfVd vEy dk K 10–5 MdksfV dk gS½ Sol. Initially degree of dissociation  = Now degree of dissociation,  = 2 = = so C1 = C  Hence we have 4 300 × 0.2 = Vf × 0.2 4 so Vf = 1200 ml Hence water added = 1200 – 300 = 900 ml Ans. 900 mL. 6. Saccharin (Ka = 2  10 ) is a weak acid represented by formula HSac. A 4  10 mole amount of 12 4 saccharin is dissolved in 200 cm3 solution of pH 3 . Assuming no change in volume , calculate the concentration of Sac  ions in the resulting solution at equilibrium. LkSdsjhu(K =21012),dnqcZyvEygSftldksHSaclw=k}kjkçnf'kZrdjrsgSA,d4104eksyek=k dsLkSdsjhu dks200cm3foy;uesa?ksyusijpH3çkIrgksrhgSAekukdhvk;ruesadksbZifjorZuughagksrkgSAlkE;koLFk ijifj.kehs foy;uesaSacvk;udhlkUnzrkdhx.kukdjksA Ans. [Sac–] = 4 × 10–12 M Sol. Calculation of [H+] and [HSac] at start 4 10 4 1000 [HSac] = 200 = 0.002 M The dissociation of HSac is as below HSac H+ + Sac– At start 0.002 0.001 0 At equi. 0.002-x 0.001 + x x  K = [H ][Sac ] [HSac] = (0.01 x)x 0.002  x = 2 ×10–12 x = 4 ×10–12 M [Sac–] = 4 ×10–12 M. 7. Find [OH–] in a solution made by dissolving 0.005 mol each of NH (K = 1.8 × 10–5) and pyridine (K =1.5×10–9) 3 b b in enough water to make up 1 L of the solution. What are the concentration of ammonium and pyridinium ions. NH (K =1.8×10–5)rFkkfijhMhu(K =1.5×10–9)izR;sdds 0.005eksydks i;kZIrek=kk esatydksfeykdj 1yhVj 3 b b foy;ucuk;kx;kgSArksfoy;uesa[OH–]dhx.kukdjksAveksfu;erFk fijhfMuh;evk;udhlkUnzrkD;kgksxh\ Ans. [OH–] = [NH+ ] = 3 × 10–4 M ; [C H N+] = 2.5 × 10–8 M. 8. What is the concentration of acetic acid which can be added to 0.5 M formic acid so that the % dissociation of neither acid is changed by the addition. Ka for acetic acid is 2 × 10-5, K for formic acid = 2.4 × 10-4. ,lhfVdvEydhlkUnzrkD;kgSftls0.5MlkUnzrkdsQkWfeZdvEyesablizdkjfeyk;ktkrkgSfdfeykusijvEy ds% fo;kstudhek=k esadksbZifjorZuughagksrkgSA,lhfVdvEydsfy,K =2×10-5gSrFk QkWfeZdvEydsfy,K =2.4×10- a a 4gSA Sol. C  = C  1 1 2 2 =  2 × 10–5 × C = 2.4 × 10–4 × 0.5 = C = 6 M. 9. Hydrazine, N2H4, can interact with water in two stages. N H (aq) + H O(l) N H + (aq) + OH–(aq) K = 8.1 × 10–7 N H + (aq) + H O (l) N H 2+ (aq) + OH–(aq) K = 9 × 10–16 2 5 2 2 6 b2 (i) What are the concentrations of OH–, N H + and N H 2+ in a 0.010 M aqueous solution of hydrazine ? 2 5 2 6 (ii) What is pH of the 0.010 M solution of hydrazine? gkbMªsthuN2H4,tydslkFknksinks aesafØ;kdjrkgSA N H (aq) + H O(l) N H + (aq) + OH–(aq) K = 8.1 × 10–7 N H + (aq) + H O (l) N H 2+ (aq) + OH–(aq) K = 9 × 10–16 2 5 2 2 6 b2 (i) 0.010Mtyh; foy;u ds gkbMªsthu esaOH–,N H + rFkkN H 2+ dhlkUnzrk D;kgS \ 2 5 2 6 (ii) gkbMªsthu ds0.010Mfoy;u dhpHD;k gS \ Ans. (i) [OH–] = 9 × 10–5 M, [N2H5+] = 9 × 10–5 ; (ii) pH = 9.96, [N2H62+] = 9 × 10–16. Integer Answer Type This section contains 3 questions. The answer to each of the questions is a single digit integer, ranging from 0 to 9. iw.k±dmÙkjizdkj bl[k.Mesa3iz'ugSaAizR;sdiz'udkmÙkj0ls9 rdiw.k ±dgSA 10. In what ratio should a 15% solution of acetic acid be mixed with a 3% solution of the acid to prepare a 12% solution(all percentages are mass/mass percentages) : 3% vEydsfoy;udslkFk15%,flfVdvEydsfoy;udksfdlvuqikresa fefJrdjukpkfg,fd12%foy;u(lHkh izfr'krnzO;eku@nzO;ekuizfr'kresagSa)culdsA Ans. 3 Sol. Let M1 (mass) of 15% solution and M2 (mass) of 3%. Resultant = 12% solution. 12(M1 + M2) = 15 × M1 + 3 × M2. 3M1 = 9M2. M1 3 M2 = 1 . 11. 2 mole PCl5, 1 mole Cl2 and 1 mole PCl3 are taken in 1 lt flask. When equilibrium is set up, PCl5 is found to be 50% dissociated into the products. Then, find KC for the reaction PCl5(g) PCl3(g) + Cl2(g). ,d1ltdsik=kesa]PCl5 ds2eksy]Cl2ds,deksyrFk PCl3ds,deksyfy;sx;sAtclkE;LFk firgksrkgS]rcik;k x;kfd]PCl5 dk50%mRikn esafo;ksftrgkstkrkgSAblizdkjvfHkfØ;kPCl5(g) PCl3(g)+Cl2(g)dsfy,KC gSA Ans. 4 Sol. PCl5 PCl3 + Cl2 Initially 2 1 1 izkjEHkesa 2 1 1 At equilibrium 2 – 1 1 + 1 1 + 1 lkE;ij 2 – 1 1 + 1 1 + 1 2  2 KC = 1 = 4 12. Consider the following reversible reactions : A + B P ; KC = 6 2B + C 2D ; KC = 4 Hence, what will be the KC for the reaction, C A + D P + 2 fuEumRØe.kh;vfHkfØ;kvksaijfopkjdhft,: A + B P ; KC = 6 2B + C 2D ; KC = 4 C rc vfHkfØ;kA+D P+ 2 ds fy, KC dk eku D;k gksxk \ Ans. 3   1  Sol. Eq. (i) + {EQ. (ii)} ×    2  So, KC = 6 × (4)– 1 / 2 = 3 DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 48 Class : XIII Course : DPP No.2 Total Marks : 42 Max. Time : 42 min. Single choice Objective ('–1' negative marking) Q.1 to Q.2 (3 marks 3 min.) [6, 6] Subjective Questions ('–1' negative marking) Q.3 to Q.11 (4 marks 4 min.) [36, 36] 1.(a) Which of the following salts is alkaline in water? fuEuesalsdkSulsyo.kdktyesa{k jh;gSA (1) CuSO4 . 5H2O (2) K2SO4 (3) NaNO3 (4*) Na2B4O7 . 10H2O (b) Which of the following salts undergoes anionic hydrolysis? fuEuesalsdkSulkyo.kdk_.kvk;fudtyvi?kVun'kkZrkgSA (1) CuSO4 (2) NH4Cl (3) FeCl3 (4*) Na2CO3 2.(a) The pH value will be highest for the aqueous solution of fuEu esals fdlyo.k dstyh; foy;udh pHdkeku vf/kdregksxk & (1) NaCl (2*) Na2CO3 (3) NH4Cl (4) NaHCO3 (b) Which of the following salts does not undergo hydrolysis? fuEuesalsdkSulsyo.kdktyvi?kVuughagksrkgSA (1) NH4NO3 (2) FeCl3 . 6H2O (3*) KCl (4) KCN 3. Find the concentration of H+ , HCO  & CO 2 in a 0.01 M solution of H CO if the pH of this is 4.1. 3 K (H CO ) = 4  10 7 ; K 3 2 3 (HCO ) = 4.8  10 11 . a 2 3 a 3 0.01MH CO ds foy;u esaH+ ,HCO  rFkkCO 2 dh lkUnzrk Kkr djksA ;fn bldhpH4.1gSA 2 3 3 3 K (H CO ) = 4  10 7 ; K (HCO ) = 4.8  10 11 . a 2 3 a 3 Ans. [H+] = 8 × 10–5, [HCO –] = 5 × 10–5, [CO 2–] = 3 × 10–11. 3 3 4.(a) Potassium Cyanide is the deadliest poison known. Calculate the percent hydrolysis in a 0.06 M solution of KCN. [K (HCN) = 6 × 10-10] iksVsf'k;elk;ukbM,de`R;qdkjhfo"kds:iesatkuktkrkgSAKCNds0.06Mfoy;udsizfr'krtyvi?kVudhx.kuk dhft;saA [K (HCN) = 6× 10-10] Sol. h = = = = 1.66 × 10–2 = 1.66 %. (b) Calculate the degree of hydrolysis of 0.1 M sodium acetate solution if pH is 8.9 . 0.1MlksfM;e,lhVsVfoy;uds tyvi?kV~u dhek=kk dhx.kuk djks;fnpH 8.9gSA Ans. 8 x 10-5. 5. A solution contains 0.10 M H S and 0.25 M HCl. Calculate the concentration of [ S 2 ] and [ HS  ] ions in the solution. For H2S, Ka1 = 1.0 ×10–7 , a = 1.3 × 10–13 ,d foy;u0.10M H S rFkk0.25M HClj[krk gSA foy;u esa [S2]rFkk[HSvk;udhlkUnzrk dh x.kuk djkssAH S 2 2 ds fy, Ka = 1.0 ×10–7 rFkk Ka = 1.3 × 10–13. Ans. [ HS  ] = 4  10 8 M ; [ S2  ] = 2.08  10 20 M. 6. Determine the [S2] in a saturated (0.1M) H S solution to which enough HCl has been added to produce a [H+] of 2 x 104 . K = 107 , K = 1014. (0.1M) H Sdslar`Irfoy;uesa[S2]vk;udhlkUnzrkdkfu/k Zj.kdhft,tksfdi;kZIrek=k esaHClfeykusij [H+] ds 2 x 104 mRiUu dj ldsA K Ans. [S2] = 2.5 x 1015. = 107 , K = 1014. Sol. Ka = [H ]2 [S2 ] (H2S) 1  10–21 = 4 10 8  [S2 ] 101 so [S2–] = × 10 –14 = 2.5 ×10–15 M. 4 7. Ka for butyric acid is 2.0  10 . Calculate pH and hydroxyl ion concentration of 0.02 M aqueous solution of 6 sodium butyrate . C;wVsfjdvEydsfy,K =2.0106gSApHrFk 0.02MlksfM;eC;wVsjsVdstyh;foy;uesagkbMªksfDlyvk;udh lkUnzrkdhx.kukdhft,A Ans. pH = 9, [OH¯ ] = 10–5. 8. Calculate the mass of NH4Cl required to be dissolved in 500 ml water to have pH = 4.5, if Kb for NH OH is 10 –6 . 500ml tyesapH=4.5,j[kus ds fy,NH Clds fdrus nzO;eku dh vko';drk gksxh\;fnNH OHdsfy,K =10–6 gSA 4 4 b Ans. 2.675 gm. 9. Calculate hydrolysis constants for each of the following salt solutions. Compute also the pH of the solution and the percentage of hydrolysis. fuEuyo.kfoy;uesaçR;sddsfy,tyvi?kV~ufLFkjkaddhx.kukdjksAfoy;udspHdhHkhx.kukdjksrFk tyvi?kV~u dkçfr'krHkhKkrdjksA (i) 0.05 M NaAc ; K (HAc) = 2 × 10–5. (ii) 0.008 M NH4 Cl ; Kb (NH ) = 2 × 10 . –5 (iii) 0.5 M Na S ; K (HS—) = 1.0 × 10–15 [ log (0.475) = - 0.32 ]. 2 a (iv) 0.64 M KCN : Ka (HCN) = 4.0 × 10–10. Ans. Kh pH %Hydrolysis Integer Answer Type This section contains 2 questions. The answer to each of the questions is a single digit integer, ranging from 0 to 9. iw.k±dmÙkjizdkj bl[k.Mesa2iz'ugSaAizR;sdiz'udkmÙkj0ls9 rdiw.k ±dgSA 10. For the equilibrium AB(g) A(g) + B(g) at a given temperature, the pressure at which one-third of AB is dissociated is numerically equal to how many times of KP ? fn;sx;srkiijlkE;AB(g) A(g) +B(g) dsfy,]nkc]ftlijABdk,dfrgkbZfo;ksftrgkstkrkgSogvkafdd :ilsKPlsfdrusxqusdscjkcjgS? Ans. 8 Sol. AB A + B a 0 0 a a a 3 3 a 4a Total moles of equilibrium = a + 3 = 3  a / 3 P a / 3 P     So, KP =  4a / 3  4a / 3  P 2a / 3 P = 8 4a / 3 or P = 8 Kp 11. How many moles of KMnO4 are needed to oxidise 50 moles of Fe0.9O in acidic medium ? vEyh;ek/;eesaFe0.9Ods50eksyksdksvkWDlhd`rdjusdsfy,KMnO4dsfdruseksyvko';dgksaxs? Ans. 7 Sol. KMnO + H+ + Fe O  Mn2+ + Fe3+ 4 0.9 vf = 5 vf = 0.9 3  20  = 0.7  nKMnO4 nFe0.9O 0.7 = 5  nKMnO4 ds vko';d eksy required = 0.7 5 × 50 = 7 moles  nKMnO4 0.7 ds vko';d eksy = 5 × 50 = 7 moles