Molarity of standard sodium hydroxide (NaOH) solution (M):0.1960
Volume (mL) of the base used to titrate of the Unknown Acid:Trial #1 13.40 - Trial #2 13.65 - Trial #3 14.40 - Trial #414.00
A. What is the Normality of each trial?
B. Calculate the average Normality of the acid for the four titrations
C. Based on the statistical evaluation of your data, calculate the Normality of the unknown acid.
2 answers
25 mL of unknown acid was used in each trial, not sure if this helps
I'm not sure how to answer this question. It is poorly worded and not all of it makes sense. Here is the best I can do under the circumstanaces.
A. The molarity of the NaOH, as stated, is 0.1960. The normality of NaOH is the same as the molarity of the NaOH solution.
B. Not sure what this means. Use mLacid x Nacid = mLbase x Nbase. For example
for trial #1, mLacid = 25.00, Nacid = ?, mL base = 13.40, Nbase = 0.1960. Do that for trial 1, trial 2, trial 3, and trial 4, then average the four final values. The average will be (sum N for the four trials/4) = ?
C. I don't know what statistical evaluation you are using.
A. The molarity of the NaOH, as stated, is 0.1960. The normality of NaOH is the same as the molarity of the NaOH solution.
B. Not sure what this means. Use mLacid x Nacid = mLbase x Nbase. For example
for trial #1, mLacid = 25.00, Nacid = ?, mL base = 13.40, Nbase = 0.1960. Do that for trial 1, trial 2, trial 3, and trial 4, then average the four final values. The average will be (sum N for the four trials/4) = ?
C. I don't know what statistical evaluation you are using.
1 answer