iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 9th April Evening Slot
In an acid-base titration, 0.1 M HCl solution
was added to the NaOH solution of unknown
strength. Which of the following correctly
shows the change of pH of the titraction
mixture in this experiment?
A.
(C)
B.
(A)
C.
(B)
D.
(D)
Correct Answer: B
Explanation:
When strong acid is added to the strong base there will be sharp in the pH. So (A) is the correct answer.
2019
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 8th April Morning Slot
If solublity product of Zr3(PO4)4 is denoted by Ksp and its molar solubility is denoted by S, then
which of the following relation between S and Ksp is correct ?
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 10th January Morning Slot
A mixture of 100 m mol of Ca(OH)2 and 2 g of sodium sulphate was dissolved in water and the volume was made up to 100 mL. The mass of calcium sulphate formed and the concentration of OH– in resulting solution, respectively, are : (Molar mass of Ca (OH)2, Na2SO4 and CaSO4 are 74, 143 and 136 g mol–1
, respectively; Ksp of Ca(OH)2 is 5.5 × 10–6
)
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2018 (Offline)
An alkali is titrated against an acid with methyl orange as indicator, which of the following is a correct combination?
A.
Base
Acid
End point
Weak
Strong
Colourless to pink
B.
Base
Acid
End point
Strong
Strong
Pinkish red to yellow
C.
Base
Acid
End point
Weak
Strong
Yellow to pinkish red
D.
Base
Acid
End point
Strong
Strong
Pink to colourless
Correct Answer: C
Explanation:
Methyl orange is a pH indicator frequently used in titrations because of its clear and distinct colour change. Under acidic conditions, it is red or pinkish red and under alkaline conditions, it turns yellow.
Let's consider each of the options :
Option A : When a weak base is titrated against a strong acid, the solution at the end point is slightly acidic. However, the color change from colorless to pink is not associated with methyl orange.
Option B : When a strong base is titrated against a strong acid, the solution is neutral at the end point. Methyl orange changes color from pinkish-red in acidic conditions to yellow in basic conditions, so this color change doesn't match the description.
Option C : When a weak base is titrated against a strong acid, the solution is slightly acidic at the end point. Methyl orange would change from yellow in the basic solution to red in the acidic solution at the end point. This corresponds to a "yellow to pinkish-red" color change.
Option D : When a strong base is titrated against a strong acid, the solution is neutral at the end point. The color change described here is "pink to colorless" which does not correspond with the characteristic changes of methyl orange.
Therefore, the correct answer is Option C: Weak base, Strong acid, Yellow to pinkish-red.
2018
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2018 (Offline)
Which of the following salts is the most basic in aqueous solution?
A.
Pb(CH3COO)2
B.
Al(CN)3
C.
CH3COOK
D.
FeCl3
Correct Answer: C
Explanation:
Note : When a salt is made with strong base and weak acid then that salt will be basic nature in aqueous solution.
(a) Pb(CH3COO)2 is a salt of weak base Pb(OH)2 and weak acid CH3COOH.
(b) Al(CN)3 is a salt of weak base Al(OH)3 and weak acid HCN.
(c) CH3COOK is a salt of weak acid CH3COOH and strong base KOH.
(d) FeCl3 is a salt of weak base Fe(OH)3 and strong acid HCl.
2018
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2018 (Offline)
An aqueous solution contains an unknown concentration of Ba2+. When 50 mL of a 1 M solution of Na2SO4 is added, BaSO4 just begins to precipitate. The final volume is 500 mL. The solubility product of BaSO4 is 1 $\times$ 10–10. What is the original concentration of Ba2+?
A.
1.0 $\times$ 10–10 M
B.
5 $\times$ 10–9 M
C.
2 $\times$ 10–9 M
D.
1.1 $\times$ 10–9 M
Correct Answer: D
Explanation:
Let initially concentration of Ba+2 = x m.
After adding 50 ml Na2SO4 in Ba+2 solution final volume becomes 500 ml.
$\therefore\,\,\,$ Initial volume of Ba+2 solution
= (500 $-$ 50) ml = 450 ml
As at the begining of precipitation, ionic product = solubility product.
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2018 (Offline)
Which of the following are Lewis acids?
A.
BCl3 and AlCl3
B.
PH3 and BCl3
C.
AlCl3 and SiCl4
D.
PH3 and SiCl4
Correct Answer: A
Explanation:
The compound which have the ability to accepted at least are lone pair electron.
Structure of BCl3 is
Here B is electron deficient atom, so it can accepted lone pair. So it is a lewis acid.
Structure of AlCl3
Here, Al also a electron deficient atom so it has vacant orbital and in that vacant orbital it can take lone pair. So it is also lewis acid.
Here in PH3 there is vacant 3d orbital but it Can't take. Lone pair in 3d orbital because P is more electro-negative than H so around P atom negative charge density is created and tendency of accepting electron decreases. So PH3 is not lewis acid.
Here octet of Si full ut in has a tendency of accepting lone pair in vacant 3d orbital.
This can be shown by following reaction.
So, here option (A) and (C) both are correct. But as SiCl4 is not as strong lewis acid as BCl3 and AlCl3, So we can say option (A) is more correct.
2018
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2018 (Offline)
An aqueous solution contains 0.10 M H2S and 0.20 M HCl. If the equilibrium constants for the formation of HS– from H2S is 1.0 $\times$ 10–7 and that of S2- from HS– ions is 1.2 $\times$ 10–13 then the concentration of S2- ions in aqueous solution is :
as K1 and K2 both are very low for this reaction so dissociation of H2S and HS$-$ will be very low so, the produced H+ from this reaction will also be very low.
So, we can say the concentration of H+ will be almost same as H+ in HCl.
$\therefore\,\,\,$ [ H+ ] = 0.2 M.
From the reaction, H2S $\rightleftharpoons\,$ 2H+ + S2$-$
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2018 (Online) 15th April Evening Slot
Following four solutions are prepared by mixing different volumes of NaOH and HCl of different concentrations, pH of which one of them will be equal to 1 ?
A.
100 mL ${M \over {10}}$ HCl + 100 mL ${M \over {10}}$ NaOH
B.
75 mL ${M \over {5}}$ HCl + 25 mL ${M \over {5}}$ NaOH
C.
60 mL ${M \over {10}}$ HCl + 40 mL ${M \over {10}}$ NaOH
D.
55 mL ${M \over {10}}$ HCl + 45 mL ${M \over {10}}$ NaOH
Correct Answer: B
Explanation:
(a) 100 mL ${M \over {10}}$ NaOH will nutalise
100 mL ${M \over {10}}$ HCl, so number extra
HCl will remain.
This will be neutral solution
$\therefore\,\,\,$ pH = 7
(b) Here 25 mL ${M \over 5}$ NaOH nutralise
25 mL ${M \over 5}$ HCl
$\therefore\,\,\,$ Extra HCl = 75 $-$ 25 = 50 mL
Total volume = 75 + 25 = 100 mL
$\therefore\,\,\,$ Milimole of HCl = ${{50} \over 5}$ = 10
$\therefore\,\,\,$ Concentration of HCl = ${{10} \over {100}}$ = 0.1
$\therefore\,\,\,$ milimole of HCl = ${{20} \over {10}}$ = 2
$\therefore\,\,\,$ Concentration of HCl = ${2 \over {100}}$ = 0.02 M
$\therefore\,\,\,$ pH = $-$ log (0.02) = 1.69
(d) HCl left = 55 $-$ 45 = 10 mL
$\therefore\,\,\,$ milimole of HCl = ${{10} \over {10}}$ = 1
$\therefore\,\,\,$ Concentration of HCl = ${{1} \over {100}}$ = 0.01 M
$\therefore\,\,\,$ pH = $-$ log (0.01) = 2
2018
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2018 (Online) 15th April Morning Slot
The minimum volume of water required to dissolve 0.1 g lead (II) chloride to get a saturated solution (Ksp of PbCl2 = 3.2 $ \times $ 10-8 atomic mass of Pb = 207 u ) is :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2017 (Online) 8th April Morning Slot
Additin of sodium hydroxide solution to a weak acid (HA)results in a buffer of pH 6. If ionition constant of HA is 10$-$5, the ratio of salt to acid concentration in the buffer solution will be :
Three reactions involving $H_2PO_4^−$ are given below :
(i) H3PO4 + H2O $\to$ H3O+ + $H_2PO_4^−$
(ii) $H_2PO_4^−$ + H2O $\to$ $HPO_4^{2−}$ + H3O+
(iii) $H_2PO_4^−$ + OH- $\to$H3PO4 + O2-
In which of the above does $H_2PO_4^−$ act as an acid?
A.
(ii) only
B.
(i) and (ii)
C.
(iii) only
D.
(i) only
Correct Answer: A
Explanation:
To determine in which reactions $ H_2PO_4^- $ acts as an acid, we need to understand the Bronsted-Lowry concept of acids and bases. According to this concept, an acid is a substance that donates a proton (H⁺) to another substance, while a base is a substance that accepts a proton.
Let's analyze each reaction :
1. $ H_3PO_4 + H_2O \to H_3O^+ + H_2PO_4^- $
In this reaction, $ H_3PO_4 $ is donating a proton to $ H_2O $, forming $ H_3O^+ $ and $ H_2PO_4^- $.
$ H_2PO_4^- $ is the product of this reaction and is not acting as an acid here.
2. $ H_2PO_4^- + H_2O \to HPO_4^{2−} + H_3O^+ $
In this reaction, $ H_2PO_4^- $ donates a proton to $ H_2O $, resulting in $ HPO_4^{2−} $ and $ H_3O^+ $.
$ H_2PO_4^- $ is acting as an acid in this reaction.
3. $ H_2PO_4^- + OH^- \to H_3PO_4 + O^{2-} $
This reaction is not correctly balanced and does not follow standard chemical reaction rules. The product $ O^{2-} $ is highly unlikely in aqueous solutions due to its reactivity.
A more correct reaction would be $ H_2PO_4^- + OH^- \to HPO_4^{2−} + H_2O $. In this corrected reaction, $ H_2PO_4^- $ is donating a proton to $ OH^- $, which makes it an acid.
Given the information, the correct option is :
Option A : (ii) only
This is because in reaction (ii), $ H_2PO_4^- $ is clearly acting as an acid by donating a proton to water. In reaction (i), $ H_2PO_4^- $ is not acting as an acid but rather is formed as a product. Reaction (iii) as written is chemically incorrect, but even in a corrected form, it would show $ H_2PO_4^- $ acting as an acid.
Solubility product of silver bromide is 5.0 $\times$ 10–13. The quantity of potassium bromide (molar mass taken as 120g of mol–1) to be added to 1 litre of 0.05 M solution of silver nitrate to start the
precipitation of AgBr is :
At 25°C, the solubility product of Mg(OH)2 is 1.0 $\times$ 10–11. At which pH, will Mg2+ ions start precipitating in the form of Mg(OH)2 from a solution of 0.001 M Mg2+ ions?
Solid Ba(NO3)2 is gradually dissolved in a 1.0 $\times$ 10-4 M Na2CO3 solution. At what concentration of Ba2+ will a precipitate begin to form ?
(Ksp for BaCO3 = 5.1 $\times$ 10−9 )
Four species are listed below
i. $HCO_3^−$
ii. $H_3O^+$
iii. $HSO_4^−$
iv. $HSO_3F$
Which one of the following is the correct sequence of their acid strength?
A.
iv < ii < iii < I
B.
ii < iii < i < iv
C.
i < iii < ii < iv
D.
iii < i < iv < ii
Correct Answer: C
Explanation:
The correct order of acidic strength of the given species in
The first and second dissociation constants of an acid H2A are 1.0 $\times$ 10−5
and 5.0 $\times$ 10−10 respectively. The overall dissociation constant of the acid will be :
In a sautrated solution of the sparingly soluble strong electrolyte AgIO3 (Molecular mass = 283) the
equilibrium which sets in is
AgIO3(s) $\leftrightharpoons$ Ag+(aq) + $IO_3^-$
If the solubility product constant Ksp of AgIO3 at a given temperature is 1.0 $\times$10−8, what is the mass of AgIO3 contained in 100 ml of its saturated solution?
The solubility product of a salt having general formula MX2, in water is: 4 $\times$ 10-12 . The
concentration of M2+ ions in the aqueous solution of the salt is :
The molar solubility (in ol L-1) of a sparingly soluble salt MX4 is "s". The corresponding solubility product is Ksp. 's' is given in term of Ksp by the relation :
Which one of the following statements is not true?
A.
pH + pOH = 14 for all aqueous solutions
B.
The pH of 1 $\times$ 10-8 M HCl is 8
C.
96,500 coulombs of electricity when passed through a CuSO4 solution deposits 1 gram equivalent of copper at the cathode
D.
The conjugate base of $H_2PO_4^-$ is $HPO_4^{2-}$
Correct Answer: B
Explanation:
$pH$ of an acidic solution should be less than $7.$ The reason is that from ${H_2}O.\left[ {{H^ + }} \right] = {10^{ - 7}}M$ which cannot be neglected in comparison to ${10^{ - 8}}M.$ The $pH$ can be calculated as.
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2026 (Online) 24th January Morning Shift
Consider two Group IV metal ions $\mathrm{X}^{2+}$ and $\mathrm{Y}^{2+}$.
A solution containing $0.01 \mathrm{M} \mathrm{X}^{2+}$ and $0.01 \mathrm{M} \mathrm{Y}^{2+}$ is saturated with $\mathrm{H}_2 \mathrm{~S}$. The pH at which the metal sulphide YS will form as a precipitate is $\_\_\_\_$ . (Nearest integer)
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2025 (Online) 7th April Evening Shift
One litre buffer solution was prepared by adding 0.10 mol each of $\mathrm{NH}_3$ and $\mathrm{NH}_4 \mathrm{Cl}$ in deionised water. The change in pH on addition of 0.05 mol of HCl to the above solution is ______________ $\times 10^{-2}$.
(Nearest integer)
Given : $\mathrm{pK}_{\mathrm{b}}$ of $\mathrm{NH}_3=4.745$ and $\log _{10} 3=0.477$
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2025 (Online) 7th April Morning Shift
The percentage dissociation of a salt $\left(\mathrm{MX}_3\right)$ solution at given temperature (van't Hoff factor $\mathrm{i}=2$ ) is ___________ %(Nearest integer)
Correct Answer: 33
Explanation:
To determine the percentage dissociation of the salt $\text{MX}_3$, consider the following dissociation reaction:
The van't Hoff factor, $\text{i}$, is given as 2. The formula relating $\text{i}$ to the degree of dissociation, $\alpha$, is:
$ \text{i} = 1 + (\text{n} - 1) \alpha $
Here, $\text{n}$ represents the total number of ions produced per formula unit of the salt. For $\text{MX}_3$, dissociation yields:
1 $\text{M}^{+3}$ ion
3 $\text{X}^{-}$ ions
Thus, $\text{n} = 4$. Substituting into the formula, we have:
$ 2 = 1 + (4 - 1) \alpha $
Simplifying, we solve for $\alpha$:
$ 2 = 1 + 3\alpha $
$ 3\alpha = 1 $
$ \alpha = \frac{1}{3} \approx 0.3333 $
As a percentage, this degree of dissociation is:
$ \alpha \times 100\% \approx 33.33\% $
Rounding to the nearest integer gives:
$ 33\% $
Therefore, the percentage dissociation of the salt $\text{MX}_3$ is approximately 33%.
2025
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2025 (Online) 4th April Evening Shift
The molar conductance of an infinitely dilute solution of ammonium chloride was found to be $185 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ and the ionic conductance of hydroxyl and chloride ions are 170 and $70 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$, respectively. If molar conductance of 0.02 M solution of ammonium hydroxide is $85.5 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$, its degree of dissociation is given by $x \times 10^{-1}$.
The value of $x$ is __________ . (Nearest integer)
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2025 (Online) 4th April Evening Shift
$x \mathrm{mg}$ of $\mathrm{Mg}(\mathrm{OH})_2($ molar mass $=58)$ is required to be dissolved in 1.0 L of water to produce a pH of 10.0 at 298 K . The value of $x$ is ________ mg. (Nearest integer)
(Given : $\mathrm{Mg}(\mathrm{OH})_2$ is assumed to dissociate completely in $\mathrm{H}_2 \mathrm{O}$ ]
Correct Answer: 3
Explanation:
To determine the mass of Mg(OH)₂ that needs to be dissolved to achieve a pH of 10.0, follow these steps:
Given:
pH = 10
Therefore, pOH = 14 - pH = 4
[OH⁻] = 10⁻⁴ M
First, calculate the number of moles of OH⁻ ions:
Number of moles of OH⁻ = 10⁻⁴
Since Mg(OH)₂ dissociates completely in water, from one mole of Mg(OH)₂, you get two moles of OH⁻:
Number of moles of Mg(OH)₂ = (10⁻⁴) / 2 = 5 × 10⁻⁵ moles
Next, calculate the mass of Mg(OH)₂:
Molar mass of Mg(OH)₂ = 58 g/mol
Convert the number of moles to mass:
Mass of Mg(OH)₂ = (5 × 10⁻⁵ moles) × (58 g/mol) = 2.9 × 10⁻³ g
Convert to milligrams: 2.9 × 10⁻³ g = 2.9 mg
Therefore, the calculated mass of Mg(OH)₂ required is approximately 2.9 mg.
2025
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2025 (Online) 4th April Morning Shift
The pH of a 0.01 M weak acid $\mathrm{HX}\left(\mathrm{K}_a=4 \times 10^{-10}\right)$ is found to be 5 . Now the acid solution is diluted with excess of water so that the pH of the solution changes to 6 . The new concentration of the diluted weak acid is given as $x \times 10^{-4} \mathrm{M}$. The value of $x$ is _________ (nearest integer)
$\begin{aligned}
&\text { On dilution let final concentration of } \mathrm{HX}=\mathrm{c} \mathrm{M}\\
&\mathrm{Hx}_{(\mathrm{aq})} \rightleftharpoons \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{X}_{(\mathrm{aq})}^{-}
\end{aligned}$
Data given is inconsistent & contradictory. This should be bonus.
2025
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2025 (Online) 24th January Evening Shift
The observed and normal molar masses of compound $\mathrm{MX}_2$ are 65.6 and 164 respectively. The percent degree of ionisation of $\mathrm{MX}_2$ is __________%. (Nearest integer)
Correct Answer: 75
Explanation:
To calculate the percent degree of ionization of the compound $\mathrm{MX}_2$, we begin by considering its dissociation process:
Therefore, the percent dissociation, or the percent degree of ionization, is:
$ \text{Percent dissociation} = 75\% $
2025
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2025 (Online) 23rd January Morning Shift
If 1 mM solution of ethylamine produces $\mathrm{pH}=9$, then the ionization constant $\left(\mathrm{K}_{\mathrm{b}}\right)$ of ethylamine is $10^{-x}$. The value of $x$ is _________ (nearest integer).
[The degree of ionization of ethylamine can be neglected with respect to unity.]
