Question

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A student was conducting a titration experiment and was asked to calculate the theoretical pH of the solution at various points in the titration curve. In this experiment, the student added 45.78 ml of a 0.50 M solution of HCI to a 200 ml beaker. The student then added 30.3 ml of deionized water. After adding the water to the solution, the mixture was stirred to ensure to ensure proper mixing of solution.. Next, the solution in the beaker was titrated with 25.98 ml of 0.50 M NaOH. Note that this is before the equivalence point. 4.

What is the molarity of the initial HCl solution of 45.78 mL of 0.50 M HCl after adding 30.3 mL of deionized water?

Answer 1

Q- What is the molarity of the initial HCl solution of 45.78 mL of 0.50 M HCl after adding 30.3 mL of deionized water?   ANS- this is an answer of this question------

DATE DATE III Molarity Equitien = M₂ V₂ Initial Miy MI = 0.50M V = 45.70 me M₂ = ? V₂ = V + V deionize H₂o V₂ = 45.70+ 30.3 [V₂ = 76.08 me Mi v = M2 V₂ 10.50 X 45.70 = M, X 76.00 M2 - 0.50 X 45.78 - 22.89 76.00 76.08 Mą = 0.3000 M Now final Ans M2 = 0.3M Thank you very much Good luck