Arithmetic error

My calculations for are arithmetically correct based on the values provided in the problem statement for .

For , the problem explicitly states: "The average volume of A used was $\text{22.50 cm}^{3}$." I used this value for $\text{V}_{A}$ in my calculations.

While the raw data in , when averaged (even with the identified errors in concordancy), might lead to a slightly different average titre (e.g., $\text{22.55 cm}^{3}$ if you average $\text{22.75 cm}^{3}$, $\text{22.50 cm}^{3}$, and $\text{22.40 cm}^{3}$), for , we must use the specific value given in its problem statement, which is $\text{22.50 cm}^{3}$.

Therefore, my calculations for are as follows:

a) (i) Concentration of A in $\text{mol/dm}^{3}$

$$\text{Molar mass of NaHSO}_{4} = \text{23} + \text{1} + \text{32} + (4 \times \text{16}) = \text{120 g/mol}$$ $$\text{C}_{A} = \frac{\text{10.6 g/dm}^{3}}{\text{120 g/mol}} = \text{0.08833 mol/dm}^{3}$$

The concentration of A is $\boxed{\text{0.0883 mol/dm}^{3}}$.

a) (ii) Concentration of B in $\text{mol/dm}^{3}$
The balanced equation is $\text{NaHSO}_{4}\text{(aq)} + \text{NaOH(aq)} \longrightarrow \text{Na}_{2}\text{SO}_{4}\text{(aq)} + \text{H}_{2}\text{O(l)}$, so the mole ratio $\text{n}_{A}:\text{n}_{B}$ is $\text{1:1}$.
Using $\text{C}_{A} = \text{0.08833 mol/dm}^{3}$, $\text{V}_{A} = \text{22.50 cm}^{3}$, and $\text{V}_{B} = \text{25.0 cm}^{3}$:

$$\text{C}_{B} = \frac{\text{C}_{A}\text{V}_{A}\text{n}_{B}}{\text{V}_{B}\text{n}_{A}} = \frac{(\text{0.08833 mol/dm}^{3}) \times (\text{22.50 cm}^{3}) \times 1}{(\text{25.0 cm}^{3}) \times 1}$$ $$\text{C}_{B} = \text{0.079497 mol/dm}^{3}$$

The concentration of B is $\boxed{\text{0.0795 mol/dm}^{3}}$.