1. If 17 mL of 0.010 M NaOH(aq) solution is needed to neutralize 50 mL of HCl(aq
ID: 966817 • Letter: 1
Question
1. If 17 mL of 0.010 M NaOH(aq) solution is needed to neutralize 50 mL of HCl(aq) of an unknown molarity, what is the molarity of the HCl(aq) in the solution that was titrated?
I know this answer is 0.0034M I need help on part ll
2. Imagine that the 50 mL HCl solution from question 1 was prepared by diluting 10 mL of a stock HCl solution with 40 mL of water. Based on your answer to question 1, what was the molarity of the initial 10 mL of HCl(aq) before it was diluted with 40 mL of water? Dilution calculations use: M1V1 = M2V2.
Explanation / Answer
1.
NaOH + HCl ----> NaCl + H2O
M1V1 = M2V2
17*0.01 = 50*M2
M2 = 0.0034 M
2.
50*0.0034 = 10*M2
M2 = 0.017 M
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