Question

calculate the ph of sulphuric acid(p=4.9g/l).Dencity of solution 1g/ml.dissosiation degree=0.95?

Answer 1

Question #30164, Chemistry, Inorganic Chemistry

Calculate the pH of sulphuric acid (p=4.9g/l). Density of solution 1g/ml. dissociation degree=0.95?

Solution

Sulphuric acid dissociates in water:

$$H_2SO_4 + H_2O \rightarrow HSO_4^{1-} + H_3O^+;$$

Find the molarity of sulphuric acid solution (we know that $m_{H_2SO_4}$ per 1 liter of the solution is 4.9g):

$$M(H_2SO_4) = \frac{m_{H_2SO_4}}{M_{H_2SO_4} \cdot V_{sol}} = \frac{4.9}{98 \cdot 1} = 0.05 \text{ mol/l};$$

where $M_{H_2SO_4} = 98$ g/mol is the sulphuric acid molar mass.

One mole of dissociated sulphuric acid gives one mole of hydronium ion. So, find hydronium ion molarity;

$$M(H_3O^+) = \alpha \cdot M(H_2SO_4) = 0.95 \cdot 0.05 = 0.0475 \text{ mol/l};$$

Where $\alpha = 0.95$ is sulphuric acid dissociation degree.

Find the pH of sulphuric acid solution:

$$pH = -lg[H_3O^+] = -lg[0.0475] = 1.323$$

Answer: the pH of sulphuric acid solution is 1.323.