94420 + Interview Questions in Chemistry in Class 12 Page 519 InterviewSolution

25901.	The Ph of blood is (approximately )
Answer» 7.4 5.2 11.3 9.6 Answer :A

Discussion

25902.	The pH of blood does not appreciably change by a small addition of an acid or a base because blood
Answer» CONTAINS serum protein which ACTS as buffer Contains iron as a part of the MOLECULE Can be easily coagulated It is body FLUID Solution :pH of BLOOD does bot change because it is a buffer solution.

Discussion

25903.	The pH of blood does not appreciably change by a small addition of acid or a base because blood
Answer» SERUM protein present in blood acts as BUFFER contains iron as a part of the molecule can be EASILY coagulated has a very HIGH pH. Solution :Thebuffersystempresent in bloodis`H_2 CO_(3)+nAHCO_3`

Discussion

25904.	The pH of basic buffer mixtures is given by pH = pK_(a) + log.(["Base"])/(["Salt"]), whereas pH of acidic buffer mixtures is given by : pH = pK_(1) + log.(["Salt"])/(["Acid"]). Addition of little acid or base although shows no change in pH for all practical purposes , but since the ratio (["Base"])/(["Salt"]) for (["Salt"])/(["Acid"]) changes, a slight decrease or increase in pH results. The amount of (NH_(4))_(2)SO_(4) to be added to 500 mL of 0.01 M NH_(4)OH solution (pK_(a) for NH_(4)^(+) is 9.26) to prepare a buffer of pH 8.26 is
Answer» 0.05 mole 0.025 mole 0.10 mole 0.005 mole Solution :`pH = pK_(a) + log.(["Base"])/(["SALT"]) rArr ["Base"] = (0.01 XX 500)/(500) = 0.01` `[NH_(4)^(+)] = (a xx 2)/(500)`, Let a millimole of `(NH_(4))_(2)SO_(4)` are added. `:. [Salt] = [NH_(4)^(+)]`. `pH = 9.26 + log [(0.01)/(2a//500)]` `8.26 = 9.26 + log.(0.01 xx 500)/(2a):. a = 25` `:.` Mole of `(NH_(4))_(2)SO_(4)` added `= 0.025`.

Discussion

25905.	The pH of basic buffer mixtures is given by pH = pK_(a) + log.(["Base"])/(["Salt"]), whereas pH of acidic buffer mixtures is given by : pH = pK_(1) + log.(["Salt"])/(["Acid"]). Addition of little acid or base although shows no change in pH for all practical purposes , but since the ratio (["Base"])/(["Salt"]) for (["Salt"])/(["Acid"]) changes, a slight decrease or increase in pH results. A solution containing 0.2 mole of dichloroacetic acid litre solution has [H^(+)]
Answer» 0.05 M 0.025 M 0.10 M 0.005 M Solution :Since `CHCl_(2).COOH` is relatively strong ACID having more `K_(a)`. `{:(CHCl_(2).COONA,rarr,CHCl_(2)COO^(-),+,Na^(+)),(,,0.1,,0.1),(CHCl_(2).COOH ,hArr,CHCl_(2)COO^(-) ,+,H^(+)),(0.2,,0,,0),((0.2 - x),,(x + 0.1),,x):}` `:. K_(a) = ([CHCl_(2)COO^(-)][H^(+)])/([CHCl_(2)COOH])` or `5 xx 10^(-2) = ([0.1 + x][x])/([0.2 - x])` `:. x = 0.05`.

Discussion

25906.	The pH of basic buffer mixtures is given by pH = pK_(a) + log.(["Base"])/(["Salt"]), whereas pH of acidic buffer mixtures is given by : pH = pK_(1) + log.(["Salt"])/(["Acid"]). Addition of little acid or base although shows no change in pH for all practical purposes , but since the ratio (["Base"])/(["Salt"]) for (["Salt"])/(["Acid"]) changes, a slight decrease or increase in pH results. The ratio of pH of solution (I) containing 1 mole of CH_(3)COONa and 1 mole of acetic acid in one litre is
Answer» `1:2` `2:1` `1:3` `3:1` SOLUTION :`{:(CH_(3)COONa+,HCl rarr,CH_(3)COOH+,NaCl,),(""1," "1,""0," "0,),(""0," "0,""1," "1,):}` `:. [CH_(3)COOH] = (1)/(2) =1`. `:. [H^(+)] = C alpha = C sqrt((K_(a))/(C)) = sqrt(K_(a).C) = sqrt(K_(a))` or `pH_(1) = -(1)/(2) LOG K_(a) = (1)/(2) pK_(a)` `{:(CH_(3)COOH+,CH_(3)COONa),(""1," "1):}` `:. pH = pK_(a) + log.(1)/(1) rArr pH_(2) = pK_(a) :. (pH_(1))/(pH_(2)) = (1)/(2)`.

Discussion

25907.	The pH of basic buffer mixtures is given by pH = pK_(a) + log.(["Base"])/(["Salt"]), whereas pH of acidic buffer mixtures is given by : pH = pK_(1) + log.(["Salt"])/(["Acid"]). Addition of little acid or base although shows no change in pH for all practical purposes , but since the ratio (["Base"])/(["Salt"]) for (["Salt"])/(["Acid"]) changes, a slight decrease or increase in pH results. The volume of 0.2 M NaOH needed to prepare a buffer of pH 4.74 with 50 ml of 0.2 M acetic acid (pK_(b) of CH_(3)COO^(-) = 9.26) is
Answer» 50 mL 25 mL 20 mL 10 mL Solution :Let V mL of NaOH be needed to give `CH_(3)COONa`. `{:(NaOH+,CH_(3)COO,rarr,CH_(3)COONa+,H_(2)O),(0.2 xx V,50 xx 0.2,,""0," "0),(-,[10-0.2V],,""0.2V,0.2 V):}` `:. pH = pK_(a) + LOG.(["Salt"])/(["ACID"]) = pK_(W) - pK_(b) + log.(["Salt"])/(["Acid"])` `= 14 - 9.26 + g.(["Salt"])/(["Acid"])` `= 14-9.26 + log.([(0.2V)/(50+V)])/([(10-0.2 V)/(50+V)])` `4.74 = 4.74 + log [(0.2V)/(10 - 0.2 V)] :. V = (10)/(0.4) = 25 mL`.`

Discussion

25908.	The pH of aqueous solution of 1M HCO ONH_(4), pK_(a) of HCO OH is 3.8 and pK_(b) " of " NH_(3) is 4.8
Answer» 6.5 4.8 3.8 8.6 Solution :`HCO ONH_(4)` is a salt of weak acid and weak base , `PH=1//2 pK_(w)+ 1//2 pK_(a)- 1//2pK_(B)` `THEREFORE pH=(1)/(2)xx14+(1)/(2)xx3.8-(1)/(2)xx4.8, pH=6.5`

Discussion

25909.	The pH of an aqueous solution of Mg(OH)_(2) is 9.0. If the solubility product of Mg(OH)_(2) is 1xx10^(-11), what is [Mg^(2+)] ?
Answer» `1XX10^(-5)` `1.0xx10^(-4)` `1xx10^(-2)` `0.1` SOLUTION :`pH=9, "" therefore [H^(+)]=10^(-9)` `[OH^(-)]=(1xx10^(-14))/(10^(-9))=10^(-5)` `Mg(OH)_(2)hArr Mg^(2+)+2OH^(-)` `K_(sp)=[Mg^(2+)][OH^(-)]^(2)` `1xx10^(-11)=[Mg^(2+)][10^(-5)]^(2)` `[Mg^(2+)]=(1xx10^(-11))/((10^(-5))^(2))=10^(-1)=0.1`

Discussion

25910.	The pH of an aqueous solution of HCl is 2. Its osmotic pressure at a temperature T would be equal to
Answer» 0.01 RT 0.02 RT RT 2 RT Answer :B

Discussion

25911.	The pH of an aqueous solution of a 0.1 M solution of a weak monoprotic acid which is 1% ionised is :
Answer» 1 2 3 11 Solution :`{:(HA,hArr,H^(+),+,A^(-)),(C(1-ALPHA),,c alpha,,c alpha):}` `[H^(+)]= c alpha = 0.1xx(1)/(100)=10^(-3)` `therefore PH = -log [H^(+)]=3`

Discussion

25912.	The pH of an aqueous solution of 1.0M ammonium formate will be (pk_a of formic acid = 3.8 and pK_a, of ammonium hydroxide = 4.8)
Answer» 5.5 7.5 6.1 6.5 Answer :D

Discussion

25913.	The pH of an aqueous solution is Zero. The solution is…….
Answer» Slightly acidic Strongly acidic Neutral basic SOLUTION :`pH=-log_10[H^+]` `THEREFORE [H^+]=10^(-pH)` `=10^0=1` `[H^+]=1 M` The solution is strongly acidic

Discussion

25914.	The pH of an aqueous solution having hydroxide ion concentration as 1 xx 10^(-5) is
Answer» 5 9 4.5 11 Solution : `[OH^(-)] = 1 xx 10^(-5)` `pOH = -LOG[OH^(-)] = 5` `pH + pOH = 14 rArr pH = 14 - 5 = 9`.

Discussion

25915.	The pH of an aqueous solution is Zero. The solution is
Answer» Slighly ACIDIC Strongly acidic Neutral Basic Solution :`PH = -log_(10)[H^(+)]` `:. [H^+] = 10^(-pH)` `=10^0 = 1` `[H^+] = 1M` The solution is strongly acidic.

Discussion

25916.	The pH of an aqueous solution at 25^(@)C is 6.1 its p(OH) is :
Answer» `12.2` `7.9` `3.9` `6.1` SOLUTION :At `25^(@)C, pH +pOH =14` `THEREFORE pOH = 14-6.1 =7.9`.

Discussion

25917.	The pH of an aqueous solution containing [H^(+)] = 3 xx 10^(-3) M is
Answer» 2.471 2.523 3 `-3` Solution :`pH = log.(1)/([H^(+)]) = log.(1)/([3 xx 10^(-3)]) = 2.523`.

Discussion

25918.	The pH of an acidic buffer solution & a basic buffer solution at 25^(@)C is :
Answer» `gt7,lt7` `lt7,gt7` `=7,=7` DEPENDENT upon `K_(a)`&K_(B)` of acid & base RESPECTIVELY. Solution :`pH=pK_(a)+log((["Salt"])/(["Acid"]))`

Discussion

25919.	The pH of a solution whose [OH^(-)]=10^(-7)M is :
Answer» 14 zero 7 `-7` ANSWER :C

Discussion

25920.	The pH of a solution prepared by mixing 2.0 ml of HCl solution of pH 3.0 and 3.0 ml of NaOH of pH 10.0 is
Answer» 2.5 3.5 5.5 6.5 Answer :B

Discussion

25921.	The pH of a solution resulting from the addition of 12.5mL of 0.1M HCl to 50mL of a solution containing 0.15M CH_(3)COOH & 0.2M CH_(3)COONa will be (Given : pK_(a) of CH_(3)COOH=4.74)
Answer» `4.74` `lt4.74` `gt4.74` `gt 9.26` Solution :`{:(,CH_(3)COONa,+,HCL,rarr,CH_(3)COOH,+,NaCl),(t=0,10,,1.25,,7.5,,-),(t=eq,8.75,,0,,8.75,,-):}` `pH=pK_(a)=4.74`

Discussion

25922.	The pH of a solution of H_2O_2is 6.0. Some chlorine gasis bubbled into this solution. Which of the following is correct?
Answer» HYDROGEN GAS is LIBERATED. The pH of resultant SOLUTION becomes 8.0. The pH of resultant solution becomes less than6.0 and oxygen gas is liberated `Cl_2O`is formed in the resultant solution. Solution :`H_2O_2 + Cl_2 to 2HCl + O_2` As HCL (a strong acid ) is produced , hence `[H^+]` of the solution increases, pH of the solution decreases.

Discussion

25923.	The pH of a solution obtained by mixing equal volumes of (N)/(10)NaOH and (N)/(20) HCl
Answer» 13.4 12.4 7.6 1.6 Solution :`N_(1)V_("base") - N_(2)V_("2 acid") = N_(R) (V_(1)+V_(2))` `(1)/(10) xx V -(1)/(20) xx V = N_(R)xx 2 xx V` `(0.05)/(2) = N_(R) rArr N_(R) = 0.025`. `[H^(+)] = 0.025 M` `PH = -log[0.025] = 1.6`.

Discussion

25924.	The pH of a solution obtained by mixing 50 mL of 2N HCI and 50 mL of 1 N NaOH is [log 5 = 0.7]
Answer» 1.7 1.3 0.7 0.3 Answer :D

Discussion

25925.	The pH of a solution obtained by mixing 50 ml of 1 N HCl and 30 ml of 1 N NaOH is [log 2.5 = 0.3979]
Answer» 3.979 0.6021 12.042 1.2042 Solution :No. of milli equivalent of `NaOH = 30 xx 1 = 30` No. of milli equivalent of `HCL = 50 xx 1 = 50` `:.` No. of milli equivalent of HCl left after TITRATION = 50 - 30 = 20. Total volume of the mixture = 50 + 30 = 80 ml i.e. 20 milli equivalent or 0.02 equivalent of HCl are PRESENT in 80 ml. `:.` 250 milli equivalent or 0.25 equivalentof HCl are presentin 1000 ml or 1 litre. i.e., `0.25 N HCl ~~ 0.25 N NaOH` (Monobasic) So, `[H^(+)] = -log_(10)[2.5 xx 10^(-1)] rArr pH = 1 - log_(10)[0.25]` `pH = -log_(10)[2.5 xx 10^(-2)rArr pH = 1 - log_(10)2.5` `pH = 1 - 0.3979 rArr pH = 0.6021`.

Discussion

25926.	The pH of a solution obtained by mixing 50 mL of 1N HCl and 30 mL of 1 N NaOH is [log 2.5 = 0.3979]
Answer» `0.979` `0.6021` `12.042` `1.2042` ANSWER :B

Discussion

25927.	The pH of a solution obtained by mixing 50 ml of 0.4 N HCl and 50 mL of 0.2 N NaOH is :
Answer» 13 12 `1.0` `2.0` ANSWER :C

Discussion

25928.	The pH of a solution obtained by mixing 50 ml of 0.4 M HCl with 50 ml of 0.2 M NaOH is
Answer» `-LOG 2` `-log 2 XX 10^(-1)` `7.0` `2.0` ANSWER :C

Discussion

25929.	The PH of a solution obtained by mixing 50 ml of 0.4 M HCl and 50 ml of 0.2 M NaoH IS :
Answer» -LOG 2 `-log XX 10^(-1)` 1 2 Answer :C

Discussion

25930.	The pH of a solution, obtained by mixing 50 ml of 0.4 HCl and 50 ml of 0.2 N NaOH, is
Answer» `1.0` `2.0` `3.0` `7.8` ANSWER :A

Discussion

25931.	The pH of a solution obtained by mixing 100mL of a solutionpH of =3 with 400mL of a solution of pH =4 is
Answer» `3- log 2.8` `7- log 2.8` `4- log 2.8` `5-log 2.8 ` SOLUTION :`N_(1)V_(1)+N_(2)V_(2)=N_(3)(100+400)` or `N_(3)=(0.1+0.04)/(500)=(0.14)/(500)=2.8 xx 10^(-4)M` `pH=-log (2.8 xx 10^(-4))=4=log 2.8`

Discussion

25932.	The pH of a solution is the negative logarithm to the base 10 of its hydrogen ion concentration in
Answer» MOLES PER LITRE Millimoles per litre Micromoles per litre Nanomoles per litre Solution :`[H^(+)]` is in moles per litre.

Discussion

25933.	The pH of a solution is increased from 3 to 6 , its H^+ ion concentration will be
Answer» reduced to half doubled reduced by 1000 TIMES increased by 1000 times. Solution :`PH =3, THEREFORE[H^+]- 10^(-3), pH = 6 ,therefore[H^+]- 10^(-6)` hence`[H^+]` is reducedby `10^3`times

Discussion

25934.	The pH of a solution is 6. Sufficient amount of acid is added to decrease the pH to 2. The increase in hydrogen ion concentration is :
Answer» 1. FOUR times 2. `1//4`times 3. hundred times 4. TEN thousand times Answer :D

Discussion

25935.	The pH of a solution is increased from 3 to 6. Its H^(+) ion concentration will be
Answer» Reduced to half Doubled Reduced by 1000 times INCREASED by 1000 times Solution :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.

Discussion

25936.	The ph of a solution is 5.0 to this solution sufficient acid is added to decreases the ph to 2.0 The increase in hydrogen ion concentration is :
Answer» 1000 times 5/2 times 100 times 5 times Answer :A

Discussion

25937.	The pH of a solution is 5. Its hydrogen ion concentration is increased 100 times. Its pH will be :
Answer» 7 3 nearly 7 does not change. Answer :B

Discussion

25938.	The pH of a solution is 2. If its pH is to be raised to 4, then the [H^(+)] of the original solution has to be
Answer» Doubled Halved Increased HUNDRED times Decreased hundred times SOLUTION :If PH of any solution is 2. Then `[H^(+)] = 10^(-2) M`. If pH of any solution is just double then pH = 4 and `[H^(+)]` will be `10^(-4)`.

Discussion

25939.	The pH of a solution is 2. Its pH is to be changed to 4. Then the H^+ concentration of original solution has to be:
Answer» Halved doubled increase by 100 times decrease by 100 times Answer :D

Discussion

25940.	The pH of a solution having [H^(+)]=10xx10^(-4) mol/litre will be
Answer» 4 2 3 1 Answer :C

Discussion

25941.	The pH of a solution containing NH_(4)OHandNH_(4)^(+) ia 9. if [NH_(4)^(+)]=0.1MandKa" of " NH_(4)^(+) " is "5xx10^(-10)then what is [NH_(4)OH]?
Answer» 0.05 M 20 M 0.5 M 2 M Answer :1

Discussion

25942.	The pH of a solution formed by mixing 40 mL of 0.10 M HCl and 10 mL of 0.45 M NaOH is:
Answer» 5 8 12 10 Answer :C

Discussion

25943.	The pH of a solution at 25^@C containing 0.10 M solution acetate and 0.03 M acetic acid is…….
Answer» 4.09 5.09 `6.10` 7.09 Solution :`pH=pK_a+log([SAL t])/([ACID])` `=4.57+log""0.10/0.03=5.09`

Discussion

25944.	The pH of a solution at 25°C containing 0.20 M sodium acetate and 0.06 M acetic acid is (pK_a for CH_3COOH = 4.74 and log 3 = 0.477)
Answer» 4.36 5.26 5.84 6.32 Answer :B

Discussion

25945.	The pH of a solution at 25^(@)C containing 0.10 m sodium acetate and 0.03 m acetic acid is (pK_(a) for CH_(3)COOH = 4.57)
Answer» 4.09 5.09 6.1 7.09 Solution :`PH = pK_(a) + log.(["salt"])/(["ACID"]) = 4.57 XX + log.(0.10)/(0.03) = 5.09`.

Discussion

25946.	The pH of a soloution containing 0.4 gm NaOH per litre is :
Answer» 2 12 10 11 Answer :B

Discussion

25947.	The pH of a soft drink is 3.82. Its hydrogen ion concentration will be
Answer» `1.96 XX 10^(-2)` mol/l `1.96 xx 10^(-3)` mol/l `1.5 xx 10^(-4)` mol/l `1.96 xx 10^(-1)` mol/l Solution :`pH = 3.82 = -log[H^(+)] :. [H^(+)] = 1.5 xx 10^(-4)` mole/litre

Discussion

25948.	The pH of a soft drink is 3.82. Its hydrogen ion concentration will be…….
Answer» `1.96 TIMES 10^-2 mol//L` `1.96 times 10^-3 mol//L` `1.5 times 10^-4 mol//L` `1.96 times 10^1 mol//L` SOLUTION :`pH=3.82=-log_10[H^+]` `THEREFORE[H^+]=1.5 times 10^-4 mol//litre`

Discussion

25949.	The pH of a simple acetate buffer is given by pH = pK_(a) + log.(["Salt"])/(["Acid"]) K_(a) of acetic acid = 1.8 xx 10^(-5) If [Salt] = [Acid] = 0.1 M, the pH of the solution would be about
Answer» 7 4.7 5.3 1.4 Solution :`["SALT"] = 0.1 M, ["Acid"] = 0.1 M` `K_(a) = 1.8 xx 10^(-5) , PH = -logK_(a) + log.(["Salt"])/(["Acid"])` `= -log 1.8 xx 10^(-5) + log.(0.1)/(0.1) = -log 1.8 xx 10^(-5)` `pH = 4.7`.

Discussion

25950.	The pH of a soft drink is 3.82. Its H^+ ion concentration will be:
Answer» `1.96xx10^(-2)` mol/lit `1.96xx10^(-3)` mol/lit `1.5xx10^(-4)` mol/lit `1.96xx10^(-1)` mol/lit Answer :C

Discussion

Explore topic-wise InterviewSolutions in .

The Ph of blood is (approximately )

The pH of blood does not appreciably change by a small addition of an acid or a base because blood

The pH of blood does not appreciably change by a small addition of acid or a base because blood

The pH of aqueous solution of 1M HCO ONH_(4), pK_(a) of HCO OH is 3.8 and pK_(b) " of " NH_(3) is 4.8

The pH of an aqueous solution of Mg(OH)_(2) is 9.0. If the solubility product of Mg(OH)_(2) is 1xx10^(-11), what is [Mg^(2+)] ?

The pH of an aqueous solution of HCl is 2. Its osmotic pressure at a temperature T would be equal to

The pH of an aqueous solution of a 0.1 M solution of a weak monoprotic acid which is 1% ionised is :

The pH of an aqueous solution of 1.0M ammonium formate will be (pk_a of formic acid = 3.8 and pK_a, of ammonium hydroxide = 4.8)

The pH of an aqueous solution is Zero. The solution is…….

The pH of an aqueous solution having hydroxide ion concentration as 1 xx 10^(-5) is

The pH of an aqueous solution is Zero. The solution is

The pH of an aqueous solution at 25^(@)C is 6.1 its p(OH) is :

The pH of an aqueous solution containing [H^(+)] = 3 xx 10^(-3) M is

The pH of an acidic buffer solution & a basic buffer solution at 25^(@)C is :

The pH of a solution whose [OH^(-)]=10^(-7)M is :

The pH of a solution prepared by mixing 2.0 ml of HCl solution of pH 3.0 and 3.0 ml of NaOH of pH 10.0 is

The pH of a solution resulting from the addition of 12.5mL of 0.1M HCl to 50mL of a solution containing 0.15M CH_(3)COOH & 0.2M CH_(3)COONa will be (Given : pK_(a) of CH_(3)COOH=4.74)

The pH of a solution of H_2O_2is 6.0. Some chlorine gasis bubbled into this solution. Which of the following is correct?

The pH of a solution obtained by mixing equal volumes of (N)/(10)NaOH and (N)/(20) HCl

The pH of a solution obtained by mixing 50 mL of 2N HCI and 50 mL of 1 N NaOH is [log 5 = 0.7]

The pH of a solution obtained by mixing 50 ml of 1 N HCl and 30 ml of 1 N NaOH is [log 2.5 = 0.3979]

The pH of a solution obtained by mixing 50 mL of 1N HCl and 30 mL of 1 N NaOH is [log 2.5 = 0.3979]

The pH of a solution obtained by mixing 50 ml of 0.4 N HCl and 50 mL of 0.2 N NaOH is :

The pH of a solution obtained by mixing 50 ml of 0.4 M HCl with 50 ml of 0.2 M NaOH is

The PH of a solution obtained by mixing 50 ml of 0.4 M HCl and 50 ml of 0.2 M NaoH IS :

The pH of a solution, obtained by mixing 50 ml of 0.4 HCl and 50 ml of 0.2 N NaOH, is

The pH of a solution obtained by mixing 100mL of a solutionpH of =3 with 400mL of a solution of pH =4 is

The pH of a solution is the negative logarithm to the base 10 of its hydrogen ion concentration in

The pH of a solution is increased from 3 to 6 , its H^+ ion concentration will be

The pH of a solution is 6. Sufficient amount of acid is added to decrease the pH to 2. The increase in hydrogen ion concentration is :

The pH of a solution is increased from 3 to 6. Its H^(+) ion concentration will be

The ph of a solution is 5.0 to this solution sufficient acid is added to decreases the ph to 2.0 The increase in hydrogen ion concentration is :

The pH of a solution is 5. Its hydrogen ion concentration is increased 100 times. Its pH will be :

The pH of a solution is 2. If its pH is to be raised to 4, then the [H^(+)] of the original solution has to be

The pH of a solution is 2. Its pH is to be changed to 4. Then the H^+ concentration of original solution has to be:

The pH of a solution having [H^(+)]=10xx10^(-4) mol/litre will be

The pH of a solution containing NH_(4)OHandNH_(4)^(+) ia 9. if [NH_(4)^(+)]=0.1MandKa" of " NH_(4)^(+) " is "5xx10^(-10)then what is [NH_(4)OH]?

The pH of a solution formed by mixing 40 mL of 0.10 M HCl and 10 mL of 0.45 M NaOH is:

The pH of a solution at 25^@C containing 0.10 M solution acetate and 0.03 M acetic acid is…….

The pH of a solution at 25°C containing 0.20 M sodium acetate and 0.06 M acetic acid is (pK_a for CH_3COOH = 4.74 and log 3 = 0.477)

The pH of a solution at 25^(@)C containing 0.10 m sodium acetate and 0.03 m acetic acid is (pK_(a) for CH_(3)COOH = 4.57)

The pH of a soloution containing 0.4 gm NaOH per litre is :

The pH of a soft drink is 3.82. Its hydrogen ion concentration will be

The pH of a soft drink is 3.82. Its hydrogen ion concentration will be…….

The pH of a simple acetate buffer is given by pH = pK_(a) + log.(["Salt"])/(["Acid"]) K_(a) of acetic acid = 1.8 xx 10^(-5) If [Salt] = [Acid] = 0.1 M, the pH of the solution would be about

The pH of a soft drink is 3.82. Its H^+ ion concentration will be: