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

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: $$\text{Volume of NaOH} = 23.60 \text{ cm}^3 - 11.90 \text{ cm}^3 = 11.70 \text{ cm}^3$$

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: $$\text{Volume of NaOH} = \frac{11.70 \text{ cm}^3}{1000 \text{ cm}^3/\text{L}} = 0.0117 \text{ L}$$

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: $$\text{Moles of KHP} = \frac{0.22000 \text{ g}}{204.24 \text{ g/mol}} = 0.00108 \text{ mol}$$

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: $$\text{Molarity of NaOH} = \frac{0.00108 \text{ mol}}{0.0117 \text{ L}} = 0.092 \text{ M}$$

The average molarity was found by summing the molarities from each trial and dividing by the number of trials.

$$\text{Average Molarity of NaOH} = \frac{0.105 \text{ M} + 0.118 \text{ M} + 0.092 \text{ M}}{3} = \frac{0.315 \text{ M}}{3} = \boxed{0.105 \text{ M}}$$