JEE Main 2019 (Online) 9th April Evening Slot | Ionic Equilibrium Question 105 | Chemistry

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?

JEE Main 2019 (Online) 9th April Evening Slot Chemistry - Ionic Equilibrium Question 105 English

(C)

(A)

(B)

(D)

JEE Main 2019 (Online) 8th April Morning Slot

MCQ (Single Correct Answer)

-1

If solublity product of Zr₃(PO₄)₄ is denoted by K_sp and its molar solubility is denoted by S, then which of the following relation between S and K_sp is correct ?

$$S = {\left( {{{{K_{sp}}} \over {929}}} \right)^{1/9}}$$

$$S = {\left( {{{{K_{sp}}} \over {6912}}} \right)^{1/7}}$$

$$S = {\left( {{{{K_{sp}}} \over {144}}} \right)^{1/6}}$$

$$S = {\left( {{{{K_{sp}}} \over {216}}} \right)^{1/7}}$$

JEE Main 2019 (Online) 12th January Evening Slot

MCQ (Single Correct Answer)

-1

If K_sp of Ag₂CO₃ is 8 $$ \times $$ 10^–12, the molar solubility of Ag₂CO₃ in 0.1 M AgNO₃ is -

8 $$ \times $$ 10^–12 M

8 $$ \times $$ 10^–10 M

8 $$ \times $$ 10^–13 M

8 $$ \times $$ 10^–11 M

JEE Main 2019 (Online) 10th January Morning Slot

MCQ (Single Correct Answer)

-1

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; K_sp of Ca(OH)₂ is 5.5 × 10^–6 )

13.6g, 0.28 mol L^$$-$$1

13.6g, 0.14 mol L^$$-$$1

1.9g, 0.28 mol L^$$-$$1

1.9g, 0.14 mol L^$$-$$1

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