4. A student obtained the following results for the standardization of their NaO

Question

4. A student obtained the following results for the standardization of their NaOH solution. Trial 1: 1.112 M Trial 2: 0.9872 M Trial 3: 0.9852 M Should the student do a fourth trial? Regardless of your answer above, the student went ahead and did a fourth trial: Trial 4: 0.9901 M What number should be used for the average concentration of the NaOH? 5. A student standardized a NaOH solution and found the average concentration to be 0.09981M. When added to 25.0 ml of an aqueous HCl solution of unknown concentration, it took exactly 24.97 mL of the standardized NaOH solution to reach the equivalence point. What is the concentration of the HCl solution? You must show your work to receive credit. Use the correct number of significant figures.

Explanation / Answer

4. All the three trails gave different concentration so student should do a fourth trail.

Value in trail 1 is not a precise value, it is very when compare with remaining three values. Remaining three are comparable. Average should be taken by using three precise values.

Average concentration = (0.9872+0.9852+0.9901)/3

= 0.9875 M

5. HCl(aq) + NaOH (aq) -------> NaCl(aq) + H2O(l)

Acid and base are reacted with 1:1 molar ratio.

At neutralization point

(Molarity of acid *volume of acid)

= molarity of base*volume of base

M*25.0mL= 0.09981M*24.97mL

M= (0.09981M*24.97mL)/25.0mL

= 0.09969M

= 0.0997M

Concentration of the HCl solution= 0.0997M

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