50 cm³ of a basic solution with a concentration of 0.05 mol dm⁻³ reacts with 22.5 cm³ of 1.0 mol dm⁻³ NaOH. Calculate the concentration of the basic solution (K<0xE2><0x82><0x99>=20).

Step 1: Amount of NaOH used

$n_{\ce{NaOH}} = c_{\ce{NaOH}} \times V_{\ce{NaOH}}$

$V_{\ce{NaOH}} = 25~\ce{cm^3} = \frac{25}{1000}~\ce{dm^3} = 0.025~\ce{dm^3}$

$n_{\ce{NaOH}} = 0.50~\ce{mol dm^{-3}} \times 0.025~\ce{dm^3} = 0.0125~\ce{mol}$

or $1.25 \times 10^{-2}~\ce{mol}$

Step 2: Amount of HCl used

$n_{\ce{HCl}} = c_{\ce{HCl}} \times V_{\ce{HCl}}$

$V_{\ce{HCl}} = 25~\ce{cm^3} = 0.025~\ce{dm^3}$

$n_{\ce{HCl}} = 0.10~\ce{mol dm^{-3}} \times 0.025~\ce{dm^3} = 0.0025~\ce{mol}$

or $2.5 \times 10^{-3}~\ce{mol}$

Final answer: Amount of NaOH used is $1.25 \times 10^{-2}~\ce{mol}$, amount of HCl used is $2.5 \times 10^{-3}~\ce{mol}$.

1.25 \times 10^{-2 mol NaOH, 2.5 \times 10^{-3} mol HCl}