A titration was performed using 25.00 cm³ of an unknown acid (H₂Y) and a 0.10 mol dm⁻³ solution of sodium hydroxide (NaOH). The average titre of NaOH used was 22.50 cm³. Calculate the concentration of

The relative atomic masses are $\text{Na}=23$, $\text{O}=16$, $\text{H}=1$.

The concentration of B is given as $4.0 \text{ g dm}^{-3}$.

The given equation is: $\text{H}_2\text{Y} + 2\text{NaOH} \rightarrow \text{Na}_2\text{Y} + 2\text{H}_2\text{O}$
From the equation, the mole ratio of acid ($\text{H}_2\text{Y}$) to base ($\text{NaOH}$) is $n_A : n_B = 1 : 2$.

Given:
Concentration of B ($C_B$) = $0.10 \text{ mol dm}^{-3}$ (from b(i))
Volume of B ($V_B$) = $25 \text{ cm}^3$
Volume of A ($V_A$) = $22.50 \text{ cm}^3$ (assumed average titre from a)

The concentration of A is given as $5.60 \text{ g dm}^{-3}$.
The concentration of A in $\text{mol dm}^{-3}$ is $0.0556 \text{ mol dm}^{-3}$ (from b(ii)).