Question

according to the equation in the first picture, I need help to fill the chart in the second picture.

sorry I filled the first picture wrong. so I update it. please use the las picture to help me fill the chart. thanks

HCl(aq) NaOHaq) NaCI(aą) H20() Part II: Concentration of HCI Using Standard NaOH Show all calculations on back of page and attach additional pages as necessary. Molarity of Standard NaOH solution: 01012ツ #1 #2 #3 . 019し50 10 . ot930L , 이920L Volume of NaOH solution in each flask (L) # Moles of NaOH in each flask # Moles of HCl to reach endpoint Volume of HCI solution to reach endpoint (L) Molarity HCI in each trial Average Molarity Deviation from Average Molarity Average Deviation Percent Average Deviation (RAD) moles nole y 0.02500レ10.02500-10.02500 Part II: Concentration of HCI Using Standard NaOH Show all calculations on back of page and attach additional pages as necessary. Molarity of Standard NOH solution: a lot 2 #1 #2 #3 6.0p. 0:01920 Volume of Nao solrion in cach flak u) 2.011 x # Moles of NaOH in each flask # Moles of HCl to reach endpoint Volume of HCI solution to reach endpoint (L) Molarity HCI in each trial Average Molarity Deviation from Average Molarity Average Deviation Percent Average Deviation (RAD) ,032 xt m eles mole s 9.02500L0.02500 o, o2so

Answer 1

1) The moles of HCl needed to reach the equivalence point are the same moles of NaOH in each flask:

1: 2.032x10 ^ -3 mol

2: 2.011x10 ^ -3 mol

3: 2.001x10 ^ -3 mol

2) The molarity of HCl of each flask is calculated:

[HCl] = n / V = ​​2.032x10 ^ -3 / 0.025 = 0.0813 M

2: 0.0804 M

3: 0.0800 M

3) Average Molarity: 0.0806 M

4) Deviation of each molarity:

D = [HCl] i - Average = 0.0813 - 0.0806 = 0.0007

2: 0.0002

3: 0.0006

5) Average deviation: 0.0015

6) The percentage of deviation is calculated:

% D = average D * 100 / Medium = 0.0015 * 100 / 0.0806 = 1.86%

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