Explain how the volume of sodium hydrocide was gotten in cm and litre and how it's molarity was gotten

Question

Explain how the volume of sodium hydrocide was gotten in cm and litre and how it's molarity was gotten

Asked on March 29, 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 · 2026-03-29T10:24:42.806Z

The volume and molarity of sodium hydroxide (NaOH) were determined through a titration experiment using potassium hydrogen phthalate (KHP) as a primary standard. Step 1: Determine the volume of NaOH used in cm³. The volume of NaOH used for each trial was calculated by subtracting the initial burette reading from the final burette reading. Trial 1: $$ \text{Volume of NaOH} = 10.20 \text{ cm}^3 - 0.00 \text{ cm}^3 = 10.20 \text{ cm}^3 $$ Trial 2: $$ \text{Volume of NaOH} = 11.90 \text{ cm}^3 - 0.00 \text{ cm}^3 = 11.90 \text{ cm}^3 $$ Trial 3: $$ \text{Volume of NaOH} = 23.60 \text{ cm}^3 - 11.90 \text{ cm}^3 = 11.70 \text{ cm}^3 $$ Step 2: Convert the volume of NaOH from cm³ to Liters (L). To convert cm³ to Liters, divide by $1000 \text{ cm}^3/\text{L}$. Trial 1: $$ \text{Volume of NaOH} = \frac{10.20 \text{ cm}^3}{1000 \text{ cm}^3/\text{L}} = 0.0102 \text{ L} $$ Trial 2: $$ \text{Volume of NaOH} = \frac{11.90 \text{ cm}^3}{1000 \text{ cm}^3/\text{L}} = 0.0119 \text{ L} $$ Trial 3: $$ \text{Volume of NaOH} = \frac{11.70 \text{ cm}^3}{1000 \text{ cm}^3/\text{L}} = 0.0117 \text{ L} $$ Step 3: Calculate the moles of KHP used in each trial. The moles of KHP were calculated using its mass and molar mass ($204.24 \text{ g/mol}$). $$ \text{Moles of KHP} = \frac{\text{Mass of KHP}}{\text{Molar mass of KHP}} $$ Trial 1: $$ \text{Moles of KHP} = \frac{0.21900 \text{ g}}{204.24 \text{ g/mol}} = 0.00107 \text{ mol} $$ Trial 2: $$ \text{Moles of KHP} = \frac{0.28900 \text{ g}}{204.24 \text{ g/mol}} = 0.00141 \text{ mol} $$ Trial 3: $$ \text{Moles of KHP} = \frac{0.22000 \text{ g}}{204.24 \text{ g/mol}} = 0.00108 \text{ mol} $$ Step 4: Determine the molarity of NaOH for each trial. Since KHP is a monoprotic acid and reacts with NaOH in a $1:1$ mole ratio, the moles of NaOH at the equivalence point are equal to the moles of KHP. $$ \text{Molarity of NaOH} = \frac{\text{Moles of NaOH}}{\text{Volume of NaOH (L)}} $$ Trial 1: $$ \text{Molarity of NaOH} = \frac{0.00107 \text{ mol}}{0.0102 \text{ L}} = 0.105 \text{ M} $$ Trial 2: $$ \text{Molarity of NaOH} = \frac{0.00141 \text{ mol}}{0.0119 \text{ L}} = 0.118 \text{ M} $$ Trial 3: $$ \text{Molarity of NaOH} = \frac{0.00108 \text{ mol}}{0.0117 \text{ L}} = 0.092 \text{ M} $$ Step 5: Calculate the average molarity of NaOH. The average molarity was found by summing the molarities from each trial and dividing by the number of trials. $$ \text{Average Molarity of NaOH} = \frac{0.105 \text{ M} + 0.118 \text{ M} + 0.092 \text{ M}}{3} = \frac{0.315 \text{ M}}{3} = \boxed{0.105 \text{ M}} $$

Accepted Answer · 2026-03-29T10:24:42.806Z

The volume and molarity of sodium hydroxide (NaOH) were determined through a titration experiment using potassium hydrogen phthalate (KHP) as a primary standard. Step 1: Determine the volume of NaOH used in cm³. The volume of NaOH used for each trial was calculated by subtracting the initial burette reading from the final burette reading. Trial 1: $$ \text{Volume of NaOH} = 10.20 \text{ cm}^3 - 0.00 \text{ cm}^3 = 10.20 \text{ cm}^3 $$ Trial 2: $$ \text{Volume of NaOH} = 11.90 \text{ cm}^3 - 0.00 \text{ cm}^3 = 11.90 \text{ cm}^3 $$ Trial 3: $$ \text{Volume of NaOH} = 23.60 \text{ cm}^3 - 11.90 \text{ cm}^3 = 11.70 \text{ cm}^3 $$ Step 2: Convert the volume of NaOH from cm³ to Liters (L). To convert cm³ to Liters, divide by $1000 \text{ cm}^3/\text{L}$. Trial 1: $$ \text{Volume of NaOH} = \frac{10.20 \text{ cm}^3}{1000 \text{ cm}^3/\text{L}} = 0.0102 \text{ L} $$ Trial 2: $$ \text{Volume of NaOH} = \frac{11.90 \text{ cm}^3}{1000 \text{ cm}^3/\text{L}} = 0.0119 \text{ L} $$ Trial 3: $$ \text{Volume of NaOH} = \frac{11.70 \text{ cm}^3}{1000 \text{ cm}^3/\text{L}} = 0.0117 \text{ L} $$ Step 3: Calculate the moles of KHP used in each trial. The moles of KHP were calculated using its mass and molar mass ($204.24 \text{ g/mol}$). $$ \text{Moles of KHP} = \frac{\text{Mass of KHP}}{\text{Molar mass of KHP}} $$ Trial 1: $$ \text{Moles of KHP} = \frac{0.21900 \text{ g}}{204.24 \text{ g/mol}} = 0.00107 \text{ mol} $$ Trial 2: $$ \text{Moles of KHP} = \frac{0.28900 \text{ g}}{204.24 \text{ g/mol}} = 0.00141 \text{ mol} $$ Trial 3: $$ \text{Moles of KHP} = \frac{0.22000 \text{ g}}{204.24 \text{ g/mol}} = 0.00108 \text{ mol} $$ Step 4: Determine the molarity of NaOH for each trial. Since KHP is a monoprotic acid and reacts with NaOH in a $1:1$ mole ratio, the moles of NaOH at the equivalence point are equal to the moles of KHP. $$ \text{Molarity of NaOH} = \frac{\text{Moles of NaOH}}{\text{Volume of NaOH (L)}} $$ Trial 1: $$ \text{Molarity of NaOH} = \frac{0.00107 \text{ mol}}{0.0102 \text{ L}} = 0.105 \text{ M} $$ Trial 2: $$ \text{Molarity of NaOH} = \frac{0.00141 \text{ mol}}{0.0119 \text{ L}} = 0.118 \text{ M} $$ Trial 3: $$ \text{Molarity of NaOH} = \frac{0.00108 \text{ mol}}{0.0117 \text{ L}} = 0.092 \text{ M} $$ Step 5: Calculate the average molarity of NaOH. The average molarity was found by summing the molarities from each trial and dividing by the number of trials. $$ \text{Average Molarity of NaOH} = \frac{0.105 \text{ M} + 0.118 \text{ M} + 0.092 \text{ M}}{3} = \frac{0.315 \text{ M}}{3} = \boxed{0.105 \text{ M}} $$