The complete neutralisation of 25.00cm of 0.200M Sodium Hydroxide solution requires 25.00cm of Sulphuric acid solution. Concentration of the acid is 0.05M. Calculate the molarity of the acid.

Molarity ya NaOH ($M_b$) = $0.200 \text{ M}$
Volume ya NaOH ($V_b$) = $25.00 \text{ cm}^3 = 0.02500 \text{ dm}^3$

$$\text{Moli za NaOH} = M_b \times V_b = 0.200 \text{ mol/dm}^3 \times 0.02500 \text{ dm}^3 = 0.00500 \text{ mol}$$

Kutoka kwa mlinganyo, moli 2 za NaOH humenyuka na moli 1 ya H₂SO₄.

$$\text{Moli za H}_2\text{SO}_4 = \text{Moli za NaOH} \times \frac{1}{2} = 0.00500 \text{ mol} \times \frac{1}{2} = 0.00250 \text{ mol}$$

Volume ya H₂SO₄ ($V_a$) = $20.00 \text{ cm}^3 = 0.02000 \text{ dm}^3$

$$\text{Molarity ya H}_2\text{SO}_4 (M_a) = \frac{\text{Moli za H}_2\text{SO}_4}{V_a} = \frac{0.00250 \text{ mol}}{0.02000 \text{ dm}^3} = 0.125 \text{ M}$$

Molarity ya asidi ni $\boxed{\text{0.125 M}}$.

Masi za atomiki: H = $1.008 \text{ g/mol}$, S = $32.07 \text{ g/mol}$, O = $16.00 \text{ g/mol}$

$$\text{Masi ya mola ya H}_2\text{SO}_4 = (2 \times 1.008) + 32.07 + (4 \times 16.00) = 2.016 + 32.07 + 64.00 = 98.086 \text{ g/mol}$$

$$\text{Concentration (g/dm}^3) = \text{Molarity (mol/dm}^3) \times \text{Masi ya mola (g/mol)}$$ $$\text{Concentration} = 0.125 \text{ mol/dm}^3 \times 98.086 \text{ g/mol} = 12.26075 \text{ g/dm}^3$$

Kuzungusha hadi tarakimu muhimu tatu:

$$\text{Concentration} = 12.3 \text{ g/dm}^3$$

Concentration ya asidi katika gdm⁻³ ni $\boxed{\text{12.3 g/dm}^3}$.