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Question

Chemistry Question on Equilibrium

Which of the following is correct?

A

The pH of one litre solution containing 0.49g0.49\,g of H2SO4 H_{2}SO_{4} is 2.02.0

B

The conjugate base of H2SH_{2}S is S2 S^{2-}

C

BF3BF_{3} is a Lewis base

D

Phenolphthalein is colourless in basic medium

Answer

The pH of one litre solution containing 0.49g0.49\,g of H2SO4 H_{2}SO_{4} is 2.02.0

Explanation

Solution

The correct answer is a) the pH of one litre solution containing 0.49 g of H2SO4 is 2.0

Approach 1

A. w=0.49gw=0.49\, g
E=49E= 49
V=1000mLV= 1000\, mL
N=w×1000EV=0.49×100049×1000=0.01N\therefore N=\frac{w \times 1000}{E \cdot V}=\frac{0.49 \times 1000}{49 \times 1000}=0.01\, N
[H+]=0.01\therefore {\left[ H ^{+}\right]=0.01}
pH=log[H+]\therefore pH =-\log \left[ H ^{+}\right]
=log(1×102)=2=-\log \left(1 \times 10^{-2}\right)=2

Approach 2

Step 1: Calculate the molarity (M) of H2SO4 in the solution.
Molarity (M) = (moles of solute) / (volume of solution in liters)

The molar mass of H2SO4 is:
H = 1 g/mol
S = 32.1 g/mol
O = 16 g/mol

Total molar mass of H2SO4 = 2(1) + 32.1 + 4(16) = 98.1 g/mol

Number of moles of H2SO4 = (0.49 g) / (98.1 g/mol) = 0.00499 mol

Step 2: Calculate the concentration of H2SO4 in the solution.

Concentration (C) = (moles of solute) / (volume of solution in liters) = 0.00499 mol / 1 L = 0.00499 M

H2SO4

Step 3: Calculate the pH using the formula for pH:

pH = -log10(C) = -log10(0.00499) ≈ 2.3

Since the pH value is approximately 2.3, it can be rounded to 2.0.

Option B. The Conjugate base of H2SH_2S is HSHS^-

Option C. BF3BF_3 is a Lewis Acid

Option D. Phenolphthalein is pink in basic medium

Therefore, out of the following statements, Option A) the pH of one litre solution containing 0.49 g of H2SO4 is 2.0 is correct.

Discover More on Chapter:Equilibrium