Part IIB. Scoring Scheme: 3-3-2-1. Calculate and enter the molarity of your three HCl standardization trials using the volume of standardized NaOH solution required for each and the average molarity of the NaOH solution from the standardization trials with KHP. You should report 3 significant figures, e. g. 0.488 M.

0.214 M.

Explanation:

The equation showing the balanced chemical reaction between HCl and NaOH is given below;

HCl + NaOH > NaCl + H​​​​​​2​​​​O

Therefore, in entry one(1) the volume of HCl solution is 10mL and the volume of NaOH solution is 21.5 mL. Hence, the molarity of HCl solution is given below;

NB: the molarity of NaOH = 0.1 M.

The molarity of HCl = molarity of NaOH × volume of NaOH/ volume of HCl solution.

The molarity of HCl = 0.1 × 21.5 / 10 = 0.215 M.

For entry two(2), the volume of HCl solution is 10mL and the volume of NaOH solution is 21.3 mL. Hence, the molarity of HCl solution is given below;

The molarity of HCl =0.1 × 21.3 / 10 = 0.213 M.

For entry three(3), the volume of HCl solution is 10mL and the volume of NaOH solution is 21.43 mL. Hence, the molarity of HCl solution is given below:

Molarity of HCl = 0.1 × 21.43 / 10 = 0.2143 M.

Therefore, let's take the average of all the molarities of HCl in the three entries.

Hence, 0.215M + 0.213M + 0.2143M / 3 = 0.214 M.

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