Homework Answers
The moles of NaOH are calculated:
n NaOH = M * V = 0.1 M * 0.0045 L = 0.00045 mol
Trial 2 = 0.0006 mol
The moles of HCl are the same as those of NaOH.
The concentration of HCl is calculated:
[HCl] = n / V = 0.00045 mol / 0.01 L = 0.045 M
Trial 2 = 0.06 M
The average HCl is calculated:
Average = 0.0525 M
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can someone please me with checking our work for the chart and
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can someone please me with checking our work for the chart and
helping me with the missing ones. Also with question 4 and 5? Thank
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I'm not sure about how I'm supposed to find NaOH
concentration from a strong acid and strong base titration. I'm
thinking that 2.5×10^-8 mol per L that I calculated would make the
moles of NaOH 2.5x10^-8 mol since it's a one to one
ratio.
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The data for the titration of 25.00 mL of Unknown Acid
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To help with your analysis we have summarized some of the key
numerical values that were determined from the original data -
using the methods on Page 13 of your lab manual to graphically
determine the equivalence point.
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