0 cm³ of 0.2 M solution of hydrochloric acid required 12.5 cm³ of sodium hydroxide solution for complete neutralisation. Calculate the concentration of sodium hydroxide.

Question

Asked on April 15, 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 · 2026-04-15T17:14:10.172Z

Here are the solutions to the problems. Question 1: 25.0 cm³ of 0.2 M solution of hydrochloric acid required 12.5 cm³ of sodium hydroxide solution for complete neutralisation. Calculate the concentration of sodium hydroxide in: (i) moles per litre. (ii) grams per litre. Step 1: Write the balanced chemical equation for the reaction. The reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH) is a neutralization reaction. HCl (aq) + NaOH (aq) NaCl (aq) + H_2O (l) The mole ratio of HCl to NaOH is 1:1. Step 2: Calculate the moles of HCl. Given: Volume of HCl (V_HCl) = 25.0 cm³ = 25.0 × 10^-3 L = 0.025 L Concentration of HCl (M_HCl) = 0.2 M (mol/L) Moles of HCl (n_HCl) = M_HCl × V_HCl n_HCl = 0.2 mol/L × 0.025 L = 0.005 mol Step 3: Calculate the moles of NaOH. From the balanced equation, 1 mole of HCl reacts with 1 mole of NaOH. So, moles of NaOH (n_NaOH) = moles of HCl = 0.005 mol. Step 4: Calculate the concentration of NaOH in moles per litre. Given: Volume of NaOH (V_NaOH) = 12.5 cm³ = 12.5 × 10^-3 L = 0.0125 L Concentration of NaOH (M_NaOH) = n_NaOHV_NaOH M_NaOH = 0.005 mol0.0125 L = 0.4 mol/L (i) The concentration of sodium hydroxide in moles per litre is: 0.4 mol/L Step 5: Calculate the concentration of NaOH in grams per litre. Molar mass of NaOH: Na = 22.99 g/mol O = 16.00 g/mol H = 1.01 g/mol Molar mass of NaOH = 22.99 + 16.00 + 1.01 = 40.00 g/mol Concentration (g/L) = Concentration (mol/L) × Molar mass (g/mol) Concentration of NaOH (g/L) = 0.4 mol/L × 40.00 g/mol = 16.0 g/L (ii) The concentration of sodium hydroxide in grams per litre is: 16.0 g/L --- Question 2: About 100 cm³ of a solution of hydrochloric acid reacted completely with 0.12g of magnesium ribbon. Calculate the concentration of the hydrochloric acid in: (i) moles per litre. (ii) grams per litre. Step 1: Write the balanced chemical equation for the reaction. The reaction between magnesium (Mg) and hydrochloric acid (HCl) produces magnesium chloride and hydrogen gas. Mg (s) + 2HCl (aq) MgCl_2 (aq) + H_2 (g) The mole ratio of Mg to HCl is 1:2. Step 2: Calculate the moles of magnesium. Given: Mass of Mg = 0.12 g Molar mass of Mg = 24.31 g/mol Moles of Mg (n_Mg) = mass of Mgmolar mass of Mg n_Mg = 0.12 g24.31 g/mol ≈ 0.004936 mol Step 3: Calculate the moles of HCl. From the balanced equation, 1 mole of Mg reacts with 2 moles of HCl. Moles of HCl (n_HCl) = 2 × n_Mg n_HCl = 2 × 0.004936 mol ≈ 0.009872 mol Step 4: Calculate the concentration of HCl in moles per litre. Given: Volume of HCl (V_HCl) = 100 cm³ = 100 × 10^-3 L = 0.100 L Concentration of HCl (M_HCl) = n_HClV_HCl M_HCl = 0.009872 mol0.100 L ≈ 0.09872 mol/L Rounding to two significant figures (based on 0.12g): M_HCl ≈ 0.099 mol/L (i) The concentration of hydrochloric acid in moles per litre is: 0.099 mol/L Step 5: Calculate the concentration of HCl in grams per litre. Molar mass of HCl: H = 1.01 g/mol Cl = 35.45 g/mol Molar mass of HCl = 1.01 + 35.45 = 36.46 g/mol Concentration (g/L) = Concentration (mol/L) × Molar mass (g/mol) Concentration of HCl (g/L) = 0.09872 mol/L × 36.46 g/mol ≈ 3.606 g/L Rounding to two significant figures: 3.6 g/L (ii) The concentration of hydrochloric acid in grams per litre is: 3.6 g/L Send me the next one 📸

Accepted Answer · 2026-04-15T17:14:10.172Z

Here are the solutions to the problems. Question 1: 25.0 cm³ of 0.2 M solution of hydrochloric acid required 12.5 cm³ of sodium hydroxide solution for complete neutralisation. Calculate the concentration of sodium hydroxide in: (i) moles per litre. (ii) grams per litre. Step 1: Write the balanced chemical equation for the reaction. The reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH) is a neutralization reaction. HCl (aq) + NaOH (aq) NaCl (aq) + H_2O (l) The mole ratio of HCl to NaOH is 1:1. Step 2: Calculate the moles of HCl. Given: Volume of HCl (V_HCl) = 25.0 cm³ = 25.0 × 10^-3 L = 0.025 L Concentration of HCl (M_HCl) = 0.2 M (mol/L) Moles of HCl (n_HCl) = M_HCl × V_HCl n_HCl = 0.2 mol/L × 0.025 L = 0.005 mol Step 3: Calculate the moles of NaOH. From the balanced equation, 1 mole of HCl reacts with 1 mole of NaOH. So, moles of NaOH (n_NaOH) = moles of HCl = 0.005 mol. Step 4: Calculate the concentration of NaOH in moles per litre. Given: Volume of NaOH (V_NaOH) = 12.5 cm³ = 12.5 × 10^-3 L = 0.0125 L Concentration of NaOH (M_NaOH) = n_NaOHV_NaOH M_NaOH = 0.005 mol0.0125 L = 0.4 mol/L (i) The concentration of sodium hydroxide in moles per litre is: 0.4 mol/L Step 5: Calculate the concentration of NaOH in grams per litre. Molar mass of NaOH: Na = 22.99 g/mol O = 16.00 g/mol H = 1.01 g/mol Molar mass of NaOH = 22.99 + 16.00 + 1.01 = 40.00 g/mol Concentration (g/L) = Concentration (mol/L) × Molar mass (g/mol) Concentration of NaOH (g/L) = 0.4 mol/L × 40.00 g/mol = 16.0 g/L (ii) The concentration of sodium hydroxide in grams per litre is: 16.0 g/L --- Question 2: About 100 cm³ of a solution of hydrochloric acid reacted completely with 0.12g of magnesium ribbon. Calculate the concentration of the hydrochloric acid in: (i) moles per litre. (ii) grams per litre. Step 1: Write the balanced chemical equation for the reaction. The reaction between magnesium (Mg) and hydrochloric acid (HCl) produces magnesium chloride and hydrogen gas. Mg (s) + 2HCl (aq) MgCl_2 (aq) + H_2 (g) The mole ratio of Mg to HCl is 1:2. Step 2: Calculate the moles of magnesium. Given: Mass of Mg = 0.12 g Molar mass of Mg = 24.31 g/mol Moles of Mg (n_Mg) = mass of Mgmolar mass of Mg n_Mg = 0.12 g24.31 g/mol ≈ 0.004936 mol Step 3: Calculate the moles of HCl. From the balanced equation, 1 mole of Mg reacts with 2 moles of HCl. Moles of HCl (n_HCl) = 2 × n_Mg n_HCl = 2 × 0.004936 mol ≈ 0.009872 mol Step 4: Calculate the concentration of HCl in moles per litre. Given: Volume of HCl (V_HCl) = 100 cm³ = 100 × 10^-3 L = 0.100 L Concentration of HCl (M_HCl) = n_HClV_HCl M_HCl = 0.009872 mol0.100 L ≈ 0.09872 mol/L Rounding to two significant figures (based on 0.12g): M_HCl ≈ 0.099 mol/L (i) The concentration of hydrochloric acid in moles per litre is: 0.099 mol/L Step 5: Calculate the concentration of HCl in grams per litre. Molar mass of HCl: H = 1.01 g/mol Cl = 35.45 g/mol Molar mass of HCl = 1.01 + 35.45 = 36.46 g/mol Concentration (g/L) = Concentration (mol/L) × Molar mass (g/mol) Concentration of HCl (g/L) = 0.09872 mol/L × 36.46 g/mol ≈ 3.606 g/L Rounding to two significant figures: 3.6 g/L (ii) The concentration of hydrochloric acid in grams per litre is: 3.6 g/L Send me the next one 📸

0 cm³ of 0.2 M solution of hydrochloric acid required 12.5 cm³ of sodium hydroxide solution for complete neutralisation. Calculate the concentration of sodium hydroxide.

Need help with your own homework?

More Chemistry Questions

0 cm³ of 0.2 M solution of hydrochloric acid required 12.5 cm³ of sodium hydroxide solution for complete neutralisation. Calculate the concentration of sodium hydroxide.

Need help with your own homework?

More Chemistry Questions