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solong [7]
3 years ago
5

Compare and contrast diffusion and osmosis. Include at least one similarity and one difference between the two processes.

Chemistry
1 answer:
Softa [21]3 years ago
4 0
Osmosis is the diffusion of water <span>across a semipermeable membrane (usually cell membrane) from a region of low solute concentration to a more concentrated solution so it can reach equilibrium (balance). 

D</span>iffusion is <span>a spontaneous movement of particles from an area of high concentration to an area of low concentration.

Both results in particles moving and help balance out the concentrations.
Also, in osmosis, the water molecules are moving. In diffusion, it is the solutes moving.

I hope this helps and explains well.</span>
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Determine the pH of a 0.100 M solution of hypobromous acid, HBrO, at 298K.
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The pH of the 0.100 M solution of hydobromous acid HBrO is 4.843

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Here we have the reaction given as follows;

HBrO ⇄ H⁺ + BrO⁻¹

Therefore;

                   HBrO ⇄ H⁺ + BrO⁻¹

Initial concentration of HBrO = 0.100 M decomposes partly to form x moles of each of H⁺ and BrO⁻¹. That is

                               HBrO          ⇄ H⁺ + BrO⁻¹

Initial                        0.1                     0       0

Change                     -x                     +x     +x

Equilibrium           0.1 - x                   x        x

Hence;

Ka =  \frac{[H^+][BrO^{-1}]}{[HBrO]} =\frac{x \times x }{0.1 - x} = \frac{x^2}{0.1 -x} = 2.06 \times 10^{-9}

(0.1 - x) × 2.06×10⁻⁹ = x²

x² + 2.06×10⁻⁹·x - 0.1 = 0

Factorizing gives;

(x + 1.4354×10⁻⁵)(x - 1.4352×10⁻⁵) = 0

Therefore, x  = 1.4352×10⁻⁵ M or -1.4354×10⁻⁵ M

We take the positive value as x is the concentration of the ions in the solution;

From the above [H⁺] = 1.4352×10⁻⁵ M

pH = -log[H⁺]  = -log(1.4352×10⁻⁵) = 4.843

The pH of the 0.100 M solution of hydobromous acid HBrO at 298 K where the Ka = 2.06×10⁻⁹ = 4.843.

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