1. A student prepared a base solution using 7 mL of 6M NaOH diluted with distill
ID: 631112 • Letter: 1
Question
1. A student prepared a base solution using 7 mL of 6M NaOH diluted with distilled water to a final volume of approximately 400 mL. The student titrated a standardized solution of hcl to determine the exact concentration of the prepared base solution. She found that it took 24.55 mL of the base solution to neutralize 20.00 mL of a 0.99897 M HCl solution. What is the molarity of the prepared naoh solution as determined in the titration? How many moles of OH are present when the end point is reached? What color should the solution appear at end point?Explanation / Answer
Using M1V1 = M2V2 M1 x 24.55 = 0.99897 x 20 Molarity of NaOH = M1 = 0.8138 M as determined by the titration. Moles of OH = molarity x volume = 0.8138 x 24.55 = 19.9794 millimoles or 19.974 x 10^-3 moles. Any acid-base indicator that changes colour between pH 4 and pH 10 is suitable to detect the end-point for a strong acid - strong base titration. Both methyl orange and phenolphthalein could be used. Just one drop of the added base will bring about a change in colour of the indicator. The colour is pink.
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