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).

Question

Asked on March 18, 2026Chemistry

This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.

Accepted Answer · 2026-03-18T11:29:14.387Z

Step 1: Amount of NaOH used n_NaOH = c_NaOH × V_NaOH V_NaOH = 25~cm^3 = (25)/(1000)~dm^3 = 0.025~dm^3 n_NaOH = 0.50~mol dm^-3 × 0.025~dm^3 = 0.0125~mol or 1.25 × 10^-2~mol Step 2: Amount of HCl used n_HCl = c_HCl × V_HCl V_HCl = 25~cm^3 = 0.025~dm^3 n_HCl = 0.10~mol dm^-3 × 0.025~dm^3 = 0.0025~mol or 2.5 × 10^-3~mol Final answer: Amount of NaOH used is 1.25 × 10^-2~mol, amount of HCl used is 2.5 × 10^-3~mol. 1.25 × 10^-2 mol NaOH, 2.5 × 10^-3 mol HCl

Accepted Answer · 2026-03-18T11:29:14.387Z

Step 1: Amount of NaOH used n_NaOH = c_NaOH × V_NaOH V_NaOH = 25~cm^3 = (25)/(1000)~dm^3 = 0.025~dm^3 n_NaOH = 0.50~mol dm^-3 × 0.025~dm^3 = 0.0125~mol or 1.25 × 10^-2~mol Step 2: Amount of HCl used n_HCl = c_HCl × V_HCl V_HCl = 25~cm^3 = 0.025~dm^3 n_HCl = 0.10~mol dm^-3 × 0.025~dm^3 = 0.0025~mol or 2.5 × 10^-3~mol Final answer: Amount of NaOH used is 1.25 × 10^-2~mol, amount of HCl used is 2.5 × 10^-3~mol. 1.25 × 10^-2 mol NaOH, 2.5 × 10^-3 mol HCl

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).

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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).

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