The concentration of NaOH in solution B is 2.00 g/dm³. Calculate the concentration of solution B in mol/dm³. If a titration used 25.00 cm³ of NaOH and 18.50 cm³ of H₂SO₄, what is the concentration of

The atomic masses are: Na = $22.99 \, \text{g/mol}$, O = $16.00 \, \text{g/mol}$, H = $1.01 \, \text{g/mol}$.

$$\text{Molar mass of NaOH} = 22.99 + 16.00 + 1.01 = 40.00 \, \text{g/mol}$$

The mass concentration of NaOH is given as $2.00 \, \text{g/dm}^3$.

$$\text{Concentration of B} = \frac{\text{Mass concentration}}{\text{Molar mass}}$$ $$\text{Concentration of B} = \frac{2.00 \, \text{g/dm}^3}{40.00 \, \text{g/mol}}$$ $$\text{Concentration of B} = 0.0500 \, \text{mol/dm}^3$$

The concentration of B is $\boxed{\text{0.0500 mol/dm}^3}$.

To calculate the pH of solution A and the amount of $\text{SO}_4^{2-}$ ions that react, the concentration of solution A ($\text{H}_2\text{SO}_4$) is required. The concentration of A is determined from the titration data (specifically, the average volume of solution A used from the burette). Since no burette readings or average volume of A are provided in the problem description, parts (ii) and (iii) cannot be calculated.

If the average volume of A ($V_a$) and the pipetted volume of B ($V_b$) were known, the calculations would proceed as follows:

• For (ii) pH of solution A:

  1. Write the balanced chemical equation: $\text{H}_2\text{SO}_4 (\text{aq}) + 2\text{NaOH} (\text{aq}) \to \text{Na}_2\text{SO}_4 (\text{aq}) + 2\text{H}_2\text{O} (\text{l})$.
  2. Use the titration formula: $\frac{C_a V_a}{n_a} = \frac{C_b V_b}{n_b}$, where $n_a = 1$ and $n_b = 2$.
  3. Calculate the concentration of A ($C_a$) using the known $C_b$ (from part i), $V_b$ (pipette volume, e.g., $25.00 \, \text{cm}^3$), and the experimental $V_a$.
  4. Since $\text{H}_2\text{SO}_4$ is a diprotic acid, $[\text{H}^+] = 2 \times C_a$.
  5. Calculate $\text{pH} = -\log_{10}[\text{H}^+]$.

• For (iii) amount of $\text{SO}_4^{2-}$ ions that react:

  1. In an acid-base neutralization reaction, $\text{SO}_4^{2-}$ ions are spectator ions and do not chemically react (i.e., their chemical form does not change).
  2. If the question implies the amount of $\text{SO}_4^{2-}$ ions present in the volume of acid that reacted, then: Moles of $\text{SO}_4^{2-}$ = Moles of $\text{H}_2\text{SO}_4$ reacted. Moles of $\text{H}_2\text{SO}_4$ reacted = $C_a \times V_a$ (where $V_a$ is the average volume of A used, in $\text{dm}^3$). This calculation also requires the concentration of A ($C_a$) and the average volume of A used ($V_a$), which are not provided.