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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.
| 25951. |
The pH ofa sample of vinegar is 3.76.Calculate the concentration of hydrogen ion in it. |
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Answer» SOLUTION :`pH=-LOG[H_3O^+]` `log[H_3O^+]=-pH=-3.76` `[H_3O^+]=ANTILOG of 4.24` `=1.738 TIMES 10^-4` `[H_3O^+]=1.74 times 10^-4 M` |
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| 25952. |
The pH of a buffer solution containing ratio of concentration of salt of weak acid and weak acid equal to 2 is (pK_(a)=4.75, log 2 = 0.3010) : |
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Answer» `4.45` or `pH=4.75+0.3010=5.05` |
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| 25953. |
The pH of a buffer solution of 0.1 M CH_(3)COOH and 0.01 M CH_(3)COONa is (pK_(a)(CH_(3)COOH)=4.745) : |
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Answer» `4.745` |
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| 25954. |
The pH of a buffer solution containing 25 ml of 1 M CH_(3)COONa and 25 ml of 1M CH_(3)COOH will be appreciably affected by 5 ml of |
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Answer» `1 M CH_(3)COOH` |
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| 25955. |
The pH of a buffer solution containing a weak acid (pK_(a)) and its salt is : |
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Answer» `PH = pK_(a)+"LOG"(["SALT"])/(["Acid"])` |
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| 25956. |
The pH of a blood stream is maintained by a proper balance of H_(2)CO_(3) and NaHCO_(3) .What volume of 5MNaHCO_(3) solution should be mixed with 10ml of a sample of solution which is 2.5M in H_(2)CO_(3) in order to maintain a pH=7.4(Take pK_(a_(1)) for H_(2)CO_(3)=6.7,log2=0.3) |
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Answer» SOLUTION :`pH=pK_(a_(1))+log""([HCO_(3)^(-)])/([H_(2)CO_(3)])` `rArr 7.4=6.7+log""([HCO_(3)^(-)])/([H_(2)CO_(3)]) rArr ([HCO_(3)^(-)])/([H_(2)CO_(3)])=5` `:.5xx"moles of" H2CO3`=Moles of `HCO_(3)^(-)` (since both are components of same solution,volume of solution WOULD be same for both) `rArr :.5xx2.5xx10xx10^(-3)=5-VrArr "required volume"=0.025L=25mL` |
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| 25957. |
The pH of a buffer containing equal molar concentrations of a weak base and its chloride (K_(b)) forweak base =2xx10^(-5), log 2=0.3) is |
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Answer» 5 `= - log K_(b)("SINCE" ("SALT")/("BASE")=1)` `=- log 2xx10^(-5)=4.7` `pH+pOH=14` `THEREFORE pH=9.3` |
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| 25958. |
The pH of a buffer solution containing 0.2 mole per litre CH_(3)COONa and 1.5 mole per litre CH_(3)COOH is (K_(a) for acetic acid is 1.8 xx 10^(-5)) |
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Answer» 4.87 `= log [1.8 xx 10^(-5)] + log.(0.2)/(1.5) = 4.87`. |
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| 25959. |
The pH of a 10^-10 M NaOH solution is nearest to: |
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Answer» 10 |
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| 25960. |
The pH of a 10^(-8) molar solution of HCl in water is |
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Answer» 8 `:.` PH = 8. But this cannot not be possible as pH of an acidic solution can not be more than 7. So we have to consider `[H^(+)]` coming from `H_(2)O`. TOTAL `[H^(+)] = [H^(+)]_(HCl) + [H^(+)]_(H_(2)O)` IONISATION of `H_(2)O : H_(2)O rarr H^(+) + OH^(-)` `K_(w) = 10^(-14) = [H^(+)][OH^(-)]` Let x be the conc. of `[H^(+)]` from `H_(2)O` or `[H^(+)] = x [OH^(-)]_(H_(2)O)` `:. 10^(-14) = (x + 10^(-8))(x)` or `x = 9.5 xx 10^(-8) M` `:. [H^(+)] = 10^(-8) + 9.5 xx 10^(-8) = 10.5 xx 10^(-8)` or pH `= -log(10.5 xx 10^(-8)) = 6.98` |
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| 25961. |
The ph of a 10^(-10) HCl solution is approximately: |
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Answer» 10 |
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| 25962. |
The pH of a 0.1 N solution of NH_4 Cl is 5.4. What will the hydrolysis constant? (supposing degree of hydrolysis as very small) |
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Answer» `2.42 xx 10^(-10)` `[H^+] =CH` `pH =-log[H^+] =-logCh = 5.4` `thereforeCh= 3.98 x 10^(-6) ` ` thereforeh=( 3.98 xx 10^(-6) )/(0.1 ) = 3.98xx 10^(-5)` Now `K_h = (h^2)/1-h^(2)= h^2 [ :. h lt LT1,therefore1- h^2~~1]` `K_h =h^2= ( 3.98 xx 10^(-5))^2 =15.8 xx 10^(10) = 1.58 xx 10^(-9)` |
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| 25963. |
The pH of a 0.1 molar solution of the acid HQ is 3. The value of the ionization constant, K_(a) of the acid is |
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Answer» `3 xx 10^(-1)` `PH = 3, [H^(+)] = 10^(-3), x = 10^(-3)` `K_(a) = ((x) xx (x))/((0.1 - x)) = ((10^(-3))^(2))/(0.1 - 10^(-3)) ~ (10^(-6))/(0.1) = 10^(-5)` |
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| 25964. |
The pH of a 0.1 molar solution of the acid HQ is 3. The value of the ionization constant, K_(a) of this acid is : |
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Answer» `1xx10^(-7)` If x is the degree of IONIZATION `0.1-x"" x ""x` Now, pH = 3 or `[H^(+)]=10^(-3) therefore x=10^(-3)` `K_(a)=(x xx x)/(0.1-x)=((10^(-3))^(2))/(0.1-10^(-3))` Assuming `0.1-10^(-3)=0.1` `K_(a)=((10^(-3))^(2))/(0.1)=1xx10^(-5)` |
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| 25965. |
The pH of a 0.1 M solution of a weak acid HA is found to be 2 at a temperature T. The osmotic pressure of the acid solution would be equal to |
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Answer» 0.11 RT |
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| 25966. |
The pH of a 0.02 M solution of hydrochloric acid is |
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Answer» `2.0` `because PH = -LOG [2 xx 10^(-2)]` pH = 1.7 i.e. in between 1 and 2. |
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| 25967. |
The pH of a 0.01 M solutions of acetic acid having degree of dissociation 12.5% is |
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Answer» 5.623 `H^(+) = 1.25 xx 10^(-3)`, PH = between 2 or 3 = 2.90. |
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| 25968. |
The pH of a 0.02 M Ca(OH)_2 solution of 25^@C is : |
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Answer» 12.6 |
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| 25969. |
The Ph of a 0.005 M aqueous solution of sulphuric acid is approximately: |
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Answer» 0.005 |
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| 25970. |
The pH of a 0.001 M NaOH will be |
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Answer» 3 `= 10^(-3) M rArr POH = 3` `pH + pOH = 14 rArr pH = 14-3 = 11`. |
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| 25971. |
The pH of 2.5xx10^(-1) M HCN solution (K_(a)=4xx10^(-10)) is : |
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Answer» `2xx10^(-5)` `= sqrt(4xx10^(-10)xx2.5xx10^(-1))` `=sqrt(1xx10^(-10))=1xx10^(-5)` `therefore PH =-log(10^(-5))=5` |
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| 25973. |
The pH of 10^(-8)M NaOH aqueous solution at 25^(@)C is |
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Answer» 7.02 |
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| 25974. |
The pH of 10^(-8) M solution of HCl in water is : |
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Answer» 8 |
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| 25975. |
The pH of 10^(-8) M HCl solution is |
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Answer» 8 From HCl, `[H^(+)]=1xx10^(-8)M` `THEREFORE`Total `[H^(+)]=(1xx10^(-7)+1xx10^(-8))M` `=(1xx10^(-7)+0.1xx10^(-7))M=1.1xx10^(-7)M` `therefore pH =-log (1.1xx10^(-7))=6.9586` |
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| 25976. |
The pH of 10^(-8) M HCl solution is : |
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Answer» 8 `THEREFORE [H^(+)]=10^(-8)M` Ionic product of water, `K_(w)=[H^(+)][OH^(+)]=10^(-14)` or `[H^(+)]=[OH^(-)]=10^(-7)M` `therefore`TOTAL `[H^(+)]=(10^(-8)+10^(-7))M` `=10^(-7)(10^(-1)+1)M` `=1.1xx10^(-7)M` `pH=-log[H^(+)]` `=-log (1.1xx10^(7))` `=6.9586`. |
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| 25977. |
The pH of 10^(-7)M NaOH is |
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Answer» 7.01 |
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| 25978. |
The pH of 10^(-5) M KOH solution will be |
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Answer» 9 `10^(-5)"M "10^(-5)"M "10^(-5)"M"` `[OH^(-)]=10^(-5)M` `pH=14-pOH` `pH=14-(-LOG[OH^(-)])` `=14+log[OH^(-)]=14+log10^(-5)` `=14-5=9` |
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| 25979. |
The pH of 10^(-7) HCl aqueous solution of HCl at 25^(@)C is : |
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Answer» `7.0` |
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| 25980. |
The pH of 10^-5 M KOH solution will be………. |
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Answer» 9 `10^-5M10^-5M10^-5M` `[OH^-]=10^-5M` `pH=14-(-LOG[OH^-])` `=14+log[OH^-]=14+log10^-5` `=14-5=9` |
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| 25981. |
The pH of 10^(-4) M KOH solution will be |
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Answer» 4 ` 10^(-4) M10^(-4) M` ` thererfore[OH^(-)] = 10^(-4) M ` ` thereforepOH=- log_(10)[OH^(-)]=- log_(10) (10^(-4)) = 4` ` therefore pH= 14-4= 10` |
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| 25982. |
The pH of 10^(-2) M Ca(OH)_(2) is |
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Answer» 12 `pOH = LOG(2 xx 10^(-2)) = 1.6990` `:. pH = 14 - 1.6990 = 12.3010`. |
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| 25985. |
The pH of 1 M solution of a weak monobasic acid (HA) is 2. Then, the van't Hoff factor is |
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Answer» `1.01` `{:(,HA,hArr,H^(+),+, A^(-)),("Initial","C MOL L"^(-1),,0,,0),("After disso.",C-C ALPHA,,Calpha,,Calpha","):}` `"Total "=1+alpha` Thus, `[H^(+)]=Calpha, i.e., 10^(-2)=1xxalphaor alpha=10^(-2)` `=1+alpha=1+0.01 =1.01` |
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| 25986. |
The ph of 1 M aqueous solution of the weak acid HA is 6.0.Find its dissociation constant. |
| Answer» Answer :B | |
| 25988. |
The pH of 0.2 M aqueous solution of NH_4CI will be (pK_a of NH_3= 4.74, log 2 = 0.3) |
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Answer» 4.98 |
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| 25989. |
The ph of 1% ionised 0.1 M solution of a weak monotrooic acid : |
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Answer» 1 |
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| 25990. |
The pH of 0.10 M NH_(3) solution (K_(b)=1.8xx10^(-5)) is : |
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Answer» `2.87` `=1.34xx10^(-3)` `[H^(+)]=(1xx10^(-14))/(1.34xx10^(-3))=7.46xx10^(-12)` `pH=-log [H^(+)]` `=-log(7.46xx10^(-12))` `=11.13` |
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| 25991. |
The pH of 0.1 M solution of the following salts increases in the order |
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Answer» `NaCl gt NH_(4)Cl LT NaCN lt HCL` NaCl neutral solution `NH_(4)Cl` slightly acidic due to the reaction `NH_(4)^(+) + H_(2)O HARR NH_(4)OH + H^(+)` NaCN slightly alkaline due to the reaction `CN^(-) + H_(2)O hArr HCN + OH^(-)` HCl highly acidic The pH of the solution will follow the order highly acidic `lt` slightly acidic `lt` neutral `lt`slightly alkaline i.e., `HCllt NH_(4)Cl lt NaCl lt NaCN`. |
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| 25992. |
The pH of 0.1 M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution. |
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Answer» Solution :`HCNO LEFTRIGHTARROW H^+ + CNO^-` pH=2.34 means `-log[H^+]=2.34 or log[H^+]=-2.34=3.86` or `[H^+]=` ANTILOG `3.86=4.57 times 10^-3 M` `[CNO^-]=[H^+]=4.57 times 10^-3M` `K_a=((4.57 times 10^-3)(4.57 times 10^-3))/0.1=2.09 times 10^-4` `a=SQRT(K_a//C)=sqrt(2.09 times 10^-4//0.1)=0.0457` |
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| 25993. |
The pH of 0.1 M solution of the following salts increases in the order: |
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Answer» Nacllt`NH_4cl`ltNaCNltHCL |
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| 25994. |
The pH of 0.1 M solution of a weak monoprotic acid 1% ionised is…… |
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Answer» 1 `=0.1 TIMES 1/100=10^-3` `[H^+]=10^-3 THEREFORE therefore pH=3` |
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| 25995. |
The pH of 0.1 M acetic acid is 3, the dissociation constant of acid will be |
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Answer» `1.0 xx 10^(-4)` `because [H^(+)] = sqrt(K xx C)` `[10^(-3)]^(2) = K xx c, ([10^(-6)])/(0.1) = K = 10^(-5)`. |
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| 25997. |
The pH of 0.05 M aqueous solution of diethylamine is 12.0. Calculate its K_b. |
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Answer» SOLUTION :HYDROLYSIS REACTION is `(C_2H_5)_2NH + H_2O = (C_2H_5)_2NH_2^(+) + OH^(-) ` `2.5 XX 10^(-3)` |
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| 25998. |
The pH of 0.01 molar solution of HCl will be |
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Answer» `0.001` |
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| 25999. |
The pH of 0.001M NaOH will be……….. |
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Answer» 3 `=10^-3 pOH=3` `pH+pOH=14` `THEREFORE pH=14-3=11` |
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| 26000. |
The pH of 0.001 M NaOH solution is |
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Answer» 3 `THEREFORE PH = -LOG (10^(-11))=11` |
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